1. A local delivery company has three packages to deliver to three different homes. if the packages are delivered at random to the three houses, how many ways are there for at least one house to get the wrong package?
a) 3
b) 5
c) 3 !
d) 5!
Discussion
Explanation: The possible outcomes that satisfy the condition of at least one house gets the wrong package are: One house gets the wrong package or two houses get the wrong package or three houses get the wrong package. We can calculate each of these cases and then add them together, or approach this problem from a different angle. The only case which is left out of the condition is the case where no wrong packages are delivered. If we determine the total number of ways the three packages can be delivered and then subtract the one case from it, the remainder will be the three cases above. There is only one way for no wrong packages delivered to occur. This is the same as everyone gets the right package. The first person must get the correct package and the second person must get the correct package and the third person must get the correct package.
⇒ 1 × 1 × 1 = 1
Determine the total number of ways the three packages can be delivered.
⇒ 3 × 2 × 1 = 6
The number of ways at least one house gets the wrong package is:
⇒ 6 - 1 = 5
Therefore there are 5 ways for at least one house to get the wrong package.
2. How many natural numbers less than a lakh can be formed with the digits 0,6 and 9?
a) 242
b) 243
c) 728
d) 729
Discussion
Explanation: The digits to be used are 0,6 and 9 The required numbers are from 1 to 99999 The numbers are five digit numbers. Therefore, every place can be filled by 0, 6 and 9 in 3 ways. Total number of ways = 3 × 3 × 3 × 3 × 3 = 35 But 00000 is also a number formed and has to be excluded. Total number of numbers, = 35 - 1
= 243 - 1
= 242
3. There are 20 couples in a party. Every person greets every person except his or her spouse. People of the same sex shake hands and those of opposite sex greet each other with a Namaste (It means bringing one's own palms together and raising them to the chest level). What is the total number of handshakes and Namaste's in the party?
a) 760
b) 1140
c) 780
d) 720
Discussion
Explanation: There are 20 men and 20 women. When a man meets a woman, there are two Namastes, whereas when a man meets a man (or a woman) there is only 1 handshake. Number of handshakes, = 2 × 20C2 (men and women ) = 2 × 190 = 380
For a number of Namastes, Every man does 19 Namastes (to the 20 women excluding his wife) and they respond in the same way. Number of Namastes
= 2 × 20 × 19
= 760 Total number of Namastes and Handshake
= 760 + 380
= 1140
4. In how many ways can a leap year have 53 Sundays
a) 365
b) 7
c) 4
d) 2
Discussion
Explanation: In a leap year there are 366 days i.e. 52 weeks + 2 extra days. So to have 53 Sundays one of these two days must be a Sunday. This can occur in only 2 ways. i.e. (Saturday and Sunday) or (Sunday and Monday). Thus number of ways = 2
5. In a certain laboratory, chemicals are identified by a colour-coding system. There are 20 different chemicals. Each one is coded with either a single colour or a unique two-colour pair. If the order of colours in the pairs does not matter, what is the minimum number of different colours needed to code all 20 chemicals with either a single colour or a unique pair of colours?
a) 7
b) 6
c) 5
d) 8
Discussion
Explanation: Each one coded with either a single colour or unique two-colour pair.
Therefore, total number of ways = n + nC2
Minimum number of different colour needed to code all 20 chemicals will be 6
= 6 + 6C2
=21
6. Out of eight crew members three particular members can sit only on the left side. Another two particular members can sit only on the right side. Find the number of ways in which the crew can be arranged so that four men can sit on each side.
a) 864
b) 863
c) 865
d) 1728
Discussion
Explanation: Required number of ways, = 3C1 × 4! × 4!
= 1728
7. A man positioned at the origin of the coordinate system. the man can take steps of unit measure in the direction North, East, West or South. Find the number of ways of he can reach the point (5,6), covering the shortest possible distance.
a) 252
b) 432
c) 462
d) 504
Discussion
Explanation: In order to reach (5,6) covering the shortest distance at the same time the man has to make 5 horizontal and 6 vertical steps. The number of ways in which these steps can be taken is given by: = $$\frac{{11!}}{{5! \times 6!}}$$
= 462
8. There are 6 equally spaced points A, B, C, D, E and F marked on a circle with radius R. How many convex pentagons of distinctly different areas can be drawn using these points as vertices ?
a) 6P5
b) 1
c) 5
d) None of these
Discussion
Explanation: Since, all the points are equally spaced; hence the area of all the convex pentagons will be same.
9. There are 2 brothers among a group of 20 persons. In how many ways can the group be arranged around a circle so that there is exactly one person between the two brothers?
a) 2 × 17!
b) 18! × 18
c) 19! × 18
d) 2 × 18!
Discussion
Explanation: n objects can be arranged around a circle in (n - 1)! If arranging these n objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number. i.e., number of arrangements = $$\frac{{\left( {n - 1} \right)!}}{2}$$ Let there be exactly one person between the two brothers as stated in the question. If we consider the two brothers and the person in between the brothers as a block, then there will 17 others and this block of three people to be arranged around a circle. The number of ways of arranging 18 objects around a circle is in 17! ways. Now the brothers can be arranged on either side of the person who is in between the brothers in 2! ways. The person who sits in between the two brothers could be any of the 18 in the group and can be selected in 18 ways. Therefore, the total number of ways
= 18 × 17! × 2
= 2 × 18!
10. a, b, c, d and e are five natural numbers. Find the number of ordered sets (a, b, c, d, e) possible such that a + b + c + d + e = 64.
a) 64C5
b) 63C4
c) 65C4
d) 63C5
Discussion
Explanation: Let assume that there are 64 identical balls which are to be arranged in 5 different compartments (Since a, b, c, d, e are distinguishable) If the balls are arranged in a row i.e., o, o, o, o, o, o . . . . (64 balls). We have 63 gaps where we can place a wall in each gap, since we need 5 compartments we need to place only 4 walls. We can do this in 63C4 ways.
11. A book-shelf can accommodate 6 books from left to right. If 10 identical books on each of the languages A,B,C and D are available, In how many ways can the book shelf be filled such that book on the same languages are not put adjacently.
a) 40P6 × 6!
b) 6P4 × 2!
c) 10 × 95
d) 4 × 35
Discussion
Explanation: First place can be filled in 4 ways. The subsequent places can be filled in 3 ways each. Hence, the number of ways
= 4 × 3 × 3 × 3 × 3 × 3
= 4 × 35
12. In how many ways can the letters of the word ABACUS be rearranged such that the vowels always appear together?
a) $$\frac{{6!}}{2}$$
b) 3! × 3!
c) $$\frac{{4!}}{2}$$
d) $$\frac{{4! \times 3!}}{{2!}}$$
Discussion
Explanation: ABACUS is a 6 letter word with 3 of the letters being vowels. If the 3 vowels have to appear together as stated in the question, then there will 3 consonants and a set of 3 vowels grouped together. One group of 3 vowels and 3 consonants are essentially 4 elements to be rearranged. The number of possible rearrangements is 4! The group of 3 vowels contains two a s and one u The 3 vowels can rearrange amongst themselves in $$\frac{{3!}}{{2!}}$$ ways as the vowel a appears twice. Hence, the total number of rearrangements in which the vowels appear together are: $$\frac{{4! \times 3!}}{{2!}}$$
13.If the letters of the word SACHIN are arranged in all possible ways and these words are written out as in dictionary, then the word SACHIN appears at serial number:
a) 601
b) 600
c) 603
d) 602
Discussion
Explanation: If the word started with the letter A then the remaining 5 positions can be filled in 5! Ways. If it started with c then the remaining 5 positions can be filled in 5! Ways. Similarly if it started with H, I, N the remaining 5 positions can be filled in 5! Ways. If it started with S then the remaining position can be filled with A, C, H, I, N in alphabetical order as on dictionary. The required word SACHIN can be obtained after the 5 × 5! = 600 Ways. i.e. SACHIN is the 601th word.
14. In how many ways can 3 men and their wives be made stand in a line such that none of the 3 men stand in a position that is ahead of his wife?
a) $$\frac{{6!}}{{3! \times 3! \times 3!}}$$
b) $$\frac{{6!}}{{2! \times 2!}}$$
c) $$\frac{{6!}}{{2! \times 2! \times 2!}}$$
d) $$\frac{{6!}}{{2! \times 3!}}$$
Discussion
Explanation: 6 people can be made to stand in a line in 6! Ways. However, the problem introduces a constraint that no man stands in a position that is ahead of his wife. For any 2 given positions out of the 6 occupied by a man and his wife, the pair cannot rearrange amongst themselves in 2! Ways as the wife has to be in a position ahead of the man. Only one of the 2! arrangements is allowed. As there are 3 couples in the group, the total number of ways gets reduced by a factor of (2! × 2! × 2!) Hence, the total number of ways, = $$\frac{{6!}}{{2! \times 2! \times 2!}}$$
15. In a cricket match if a batsman score 0, 1, 2, 3, 4 or 6 runs of a ball, then find the number or different sequences in which he can score exactly 30 runs of an over. Assume that an over consists of only 6 balls and there were no extra and no run outs
a) 86
b) 71
c) 56
d) 65
Discussion
Explanation: Case A: Five 6 and one 'zero' = $$\frac{{6!}}{{5!}}$$ = 6 Case B: Four 6 and one '2' and one '4' = $$\frac{{6!}}{{4!}}$$ = 30 Case C: Three 6 and three '4' = $$\frac{{6!}}{{3! \times 3!}}$$ = 20 Case D: Four 6 and two '3' = $$\frac{{6!}}{{4! \times 2!}}$$ = 15 Total number of different sequences = 71
16. The number of ways of arranging n students in a row such that no two boys sit together and no two girls sit together is m(m > 100). If one more student is added, then number of ways of arranging as above increases by 200%. The value of n is:
a) 12
b) 8
c) 9
d) 10
Discussion
Explanation: If n is even, then the number of boys should be equal to number of girls, let each be a. ⇒ n = 2a Then the number of arrangements = 2 × a! × a! If one more students is added, then number of arrangements, = a! × (a + 1)! But this is 200% more than the earlier ⇒ 3 × (2 × a! × a!) = a! × (a + 1)! ⇒ a + 1 = 6 and a = 5 ⇒ n = 10 But if n is odd, then number of arrangements, = a!(a + 1)! Where, n = 2a + 1 When one student is included, number of arrangements,
= 2(a + 1)! (a + 1)! By the given condition, 2(a + 1) = 3, which is not possible.
17. How many integers, greater than 999 but not greater than 4000, can be formed with the digits 0, 1, 2, 3 and 4 if repetition of digits is allowed?
a) 499
b) 500
c) 375
d) 376
Discussion
Explanation: The smallest number in the series is 1000, a 4-digit number. The largest number in the series is 4000, the only 4-digit number to start with 4. The left most digit (thousands place) of each of the 4 digit numbers other than 4000 can take one of the 3 values 1 or 2 or 3. The next 3 digits (hundreds, tens and units place) can take any of the 5 values 0 or 1 or 2 or 3 or 4. Hence, there are 3 × 5 × 5 × 5 or 375 numbers from 1000 to 3999 Including 4000, there will be 376 such numbers.
18. How many five digit positive integers that are divisible by 3 can be formed using the digits 0, 1, 2, 3, 4 and 5, without any of the digits getting repeated.
a) 15
b) 96
c) 216
d) 120
Discussion
Explanation: Test of divisibility for 3: The sum of the digits of any number that is divisible by 3 is divisible by 3 For instance, take the number 54372 Sum of its digits is 5 + 4 + 3 + 7 + 2 = 21 As 21 is divisible by 3, 54372 is also divisible by 3 There are six digits viz., 0, 1, 2, 3, 4 and 5. To form 5-digit numbers we need exactly 5 digits. So we should not be using one of the digits.
The sum of all the six digits 0, 1, 2, 3, 4 and 5 is 15. We know that any number is divisible by 3 if and only if the sum of its digits is divisible by 3
Combining the two criteria that we use only 5 of the 6 digits and pick them in such a way that the sum is divisible by 3, we should not use either 0 or 3 while forming the five digit numbers.
Case 1
If we do not use '0', then the remaining 5 digits can be arranged in:
5! ways = 120 numbers.
Case 2
If we do not use '3', then the arrangements should take into account that '0' cannot be the first digit as a 5-digit number will not start with '0'.
The first digit from the left can be any of the 4 digits 1, 2, 4 or 5
Then the remaining 4 digits including '0' can be arranged in the other 4 places in 4! ways.
So, there will be 4 × 4! numbers = 4 × 24 = 96 numbers.
Combining Case 1 and Case 2, there are a total of 120 + 96 = 216, 5 digit numbers divisible by '3' that can be formed using the digits 0 to 5.
19. There are 10 seats around a circular table. If 8 men and 2 women have to seated around a circular table, such that no two women have to be separated by at least one man. If P and Q denote the respective number of ways of seating these people around a table when seats are numbered and unnumbered, then P : Q equals
a) 9 : 1
b) 72 : 1
c) 10 : 1
d) 8 : 1
Discussion
Explanation: Initially we look at the general case of the seats not numbered. The total number of cases of arranging 8 men and 2 women, so that women are together, ⇒ 8! ×2! The number of cases where in the women are not together, ⇒ 9! - (8! × 2!) = Q Now, when the seats are numbered, it can be considered to a linear arrangement and the number of ways of arranging the group such that no two women are together is, ⇒ 10! - (9! × 2!) But the arrangements where in the women occupy the first and the tenth chairs are not favorable as when the chairs which are assumed to be arranged in a row are arranged in a circle, the two women would be sitting next to each other. The number of ways the women can occupy the first and the tenth position, = 8! × 2! The value of P = 10! - (9! × 2!) - (8! × 2!)
Thus P : Q = 10 : 1
20. How many factors of 25 × 36 × 52 are perfect squares?
a) 20
b) 24
c) 30
d) 36
Discussion
Explanation: Any factor of this number should be of the form 2a × 3b × 5c
For the factor to be a perfect square a, b, c have to be even.
a can take values 0, 2, 4, b can take values 0, 2, 4, 6 and c can take values 0, 2
Total number of perfect squares =3 × 4 × 2 = 24
21. There are five cards lying on the table in one row. Five numbers from among 1 to 100 have to be written on them, one number per card, such that the difference between the numbers on any two adjacent cards is not divisible by 4. The remainder when each of the 5 numbers is divided by 4 is written down on another card (the 6th card) in order. How many sequences can be written down on the 6th card?
a) 210
b) 210 × 33
c) 4 × 34
d) 42 × 33
Discussion
Explanation: The remainder on the first card can be 0, 1, 2 or 3 i.e. 4 possibilities. The remainder of the number on the next card when divided by 4 can have 3 possible values (except the one occurred earlier). For each value on the card the remainder can have 3 possible values. The total number of possible sequences is: 4 × 34
22. From a total of six men and four ladies a committee of three is to be formed. If Mrs. X is not willing to join the committee in which Mr. Y is a member, whereas Mr.Y is willing to join the committee only if Mrs Z is included, how many such committee are possible?
a) 138
b) 128
c) 112
d) 91
Discussion
Explanation: We first count the number of committee in which (i). Mr. Y is a member (ii). the ones in which he is not
case (i): As Mr. Y agrees to be in committee only where Mrs. Z is a member. Now we are left with (6 - 1) men and (4 - 2) ladies (Mrs. X is not willing to join). We can choose 1 more in 5+2C1 = 7 ways.
case (ii): If Mr. Y is not a member then we left with (6 + 4 - 1) people. we can select 3 from 9 in 9C3 = 84 ways. Thus, total number of ways is 7 + 84 = 91 ways.
23. Goldenrod and No Hope are in a horse race with 6 contestants. How many different arrangements of finishes are there if No Hope always finishes before Goldenrod and if all of the horses finish the race ?
a) 700
b) 360
c) 120
d) 24
Discussion
Explanation: Two horses A and B, in a race of 6 horses . . . A has to finish before B. If A finishes 1 . . . . . B could be in any of other 5 positions in 5 ways and other horses finish in 4! Ways, so total ways = 5 × 4! If A finishes 2 . . . . . B could be in any of the last 4 positions in 4 ways. But the other positions could be filled in 4! ways, so the total ways = 4 × 4! If A finishes 3rd . . . . . B could be in any of last 3 positions in 3 ways, but the other positions could be filled in 4! ways, so total ways = 3 × 4! If A finishes 4th . . . . . B could be in any of last 2 positions in 2 ways, but the other positions could be filled in 4! ways, so total ways = 2 × 4! If A finishes 5th . . . . B has to be 6th and the top 4 positions could be filled in 4! ways. A cannot finish 6th, since he has to be ahead of B. Therefore total number of ways:
= 5 × 4! + 4 × 4! + 3 × 4! + 2 × 4! + 4!
= 120 + 96 + 72 + 48 + 24
= 360
24. Jay wants to buy a total of 100 plants using exactly a sum of Rs. 1000. He can buy Rose plants at Rs. 20 per plant or marigold or Sun flower plants at Rs. 5 and Rs. 1 per plant respectively. If he has to buy at least one of each plant and cannot buy any other type of plants, then in how many distinct ways can Jay make his purchase ?
a) 2
b) 3
c) 4
d) 5
Discussion
Explanation: Let the number of Rose plants be a. Let number of marigold plants be b. Let the number of Sunflower plants be c. According to question, 20a + 5b + 1c = 1000 - - - - - - (1) a + b + c = 100 - - - - - - - - - - (2) Solving the above two equations by eliminating c, 19a + 4b = 900 b = $$\frac{{900 - 19a}}{4}$$ = $$225 - \frac{{19a}}{4}$$ - - - - - - - (3) b being the number of plants, is a positive integer, and is less than 99, as each of the other two types have at least one plant in the combination i.e . 0 < b < 99 - - - - - - - (4) Substituting (3) in (4), 0 < 225 - $$\frac{{19a}}{4}$$ < 99 ⇒ 225 < -$$\frac{{19a}}{4}$$ < (99 - 225) ⇒ 4 × 225 > 19a > 126 × 4 ⇒ $$\frac{{900}}{{19}}$$ > a > 504 a is the integer between 47 and 27 - - - - - - - - (5) From (3), it is clear, a should be multiple of 4.
Hence, possible values of a are (28,32,36,40,44)
For a=28 and 32, a+b>100
For all other values of a, we get the desired solution:
a=36,b=54,c=10
a=40,b=35,c=25
a=44,b=16,c=40
Three solutions are possible.
25. How many words of 4 consonants and 3 vowels can be made from 12 consonants and 4 vowels, if all the letters are different?
a) 16C7 × 7!
b) 12C4 × 4C3 × 7!
c) 12C3 × 4C4
d) 11C4 × 4C3
Discussion
Explanation: 4 consonants out of 12 can be selected in,12C4 ways. 3 vowels can be selected in 4C3 ways. Therefore, total number of groups each containing 4 consonants and 3 vowels, = 12C4 × 4C3 Each group contains 7 letters, which can be arranging in 7! ways. Therefore required number of words, = 12C4 × 4C3 × 7!
26. In how many different ways can the letters of the word AWARE be arranged?
a) 48
b) 60
c) 120
d) 160
Discussion
Explanation: The given word contains 5 letters of which A is taken 2 times.
∴ Required number of ways
$$\eqalign{ & = \frac{{5!}}{{2!}} \cr & = \frac{{5 \times 4 \times 3 \times 2 \times 1}}{2} \cr & = 60 \cr} $$
27. A committee of 5 members is to be formed out of 3 trainees, 4 professors and 6 research associate. In how many different ways can this be done if the committee should have 2 trainees and 3 research associates?
a) 15
b) 45
c) 60
d) 9
Discussion
Explanation: Required number of ways
$$\eqalign{ & = \left( {{}^3{C_2} \times {}^6{C_3}} \right) \cr & = \left( {{}^3{C_1} \times {}^6{C_3}} \right) \cr & = \left( {3 \times \frac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}}} \right) \cr & = 60 \cr} $$
28. In how many different ways can the letters of the word CREAM be arranged?
a) 25
b) 120
c) 260
d) 480
Discussion
Explanation: The given word contains 5 letters, all different.
∴ Required number of ways
$$\eqalign{ & = 5! \cr & = \left( {5 \times 4 \times 3 \times 2 \times 1} \right) \cr & = 120 \cr} $$
29. In how many different ways can the letters of the word ALLAHABAD be arranged ?
a) 3780
b) 1890
c) 7560
d) 2520
Discussion
Explanation: The given word contains 9 letters, namely 4A, 2L, 1H, 1B and 1D.
∴ Required number of ways
$$\eqalign{ & = \frac{{9!}}{{4! \,2! \,1! \,1! \,1!}} \cr & = \frac{{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{4 \times 3 \times 2 \times 1 \times 2}} \cr & = 7560 \cr} $$
30. From a group of 7 men 6 women, 5 persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
a) 564
b) 645
c) 735
d) 756
Discussion
Explanation: Required number of ways
$$ = \left( {{}^7{C_3} \times {}^6{C_2}} \right) + $$ $$\left( {{}^7{C_4} \times {}^6{C_1}} \right) + $$ $$\left( {{}^7{C_5} \times {}^6{C_0}} \right)$$
$$ = \left\{ {\frac{{7 \times 6 \times 5}}{{3!}} \times \frac{{6 \times 5}}{{2!}}} \right\}$$ $$ + \left( {{}^7{C_3} \times {}^6{C_1}} \right)$$ $$ + \left( {{}^7{C_2} \times 1} \right)$$
$$ = \left\{ {\frac{{7 \times 6 \times 5}}{6} \times \frac{{6 \times 5}}{{2 \times 1}}} \right\}$$ $$ + \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times 6} \right)$$ $$ + \left( {\frac{{7 \times 6}}{{2 \times 1}} \times 1} \right)$$
$$\eqalign{ & = \left( {525 + 210 + 21} \right) \cr & = 756 \cr} $$
31. In how many different ways can the letters of the word ‘TRANSPIRATION’ be arranged so that the vowels always come together?
a) 2429500
b) 1360800
c) 1627800
d) None of these
Discussion
Explanation: The word ‘TRANSPIRATION’ has 13 letters in which each of T, R, A, N and I has come two times
We have to arrange TT RR NN PS (AA II O)
There are five vowels in the given words.
∴ We consider these give vowels as one letter.
∴ Required number of arrangements
$$\eqalign{ & = \frac{{9! \times 5!}}{{2!\, 2! \,2! \,2! \,2!}} \cr & = \frac{{9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 5 \times 4 \times 3 \times 2}}{{2 \times 2 \times 2 \times 2 \times 2}} \cr & = 1360800 \cr} $$
32. In how many different ways can the letters of the word CAPITAL be arranged so that the vowels always come together?
a) 120
b) 360
c) 720
d) 840
Discussion
Explanation: Keeping the vowels (AIA) together, we have CPTL (AIA).
We treat (AIA) as 1 letter.
Thus, we have to arrange 5 letters.
These can be arranged in 5! = (5 × 4 × 3 × 2 × 1) ways = 120 ways
Now, (AIA) are 3 letters with 2A and 1I
These can be arranged among themselves in
$$\frac{{3!}}{{2!}} = \frac{{3 \times 2 \times 1}}{{2 \times 1}} = 3$$ ways
∴ Required number of ways = 120 × 3 = 360
33. A committee of 5 members is to be formed by selecting out of 4 men and 5 women. In how many different ways the committee can be formed if it should have 2 men and 3 women?
a) 16
b) 36
c) 45
d) 60
Discussion
Explanation: Required number of ways
$$\eqalign{ & = {{}^4{C_2} \times {}^5{C_3}} \cr & = {{}^4{C_2} \times {}^5{C_2}} \cr & = {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \cr & = 60 \cr} $$
34. In how many different ways can the letters of the word INCREASE be arranged?
a) 40320
b) 10080
c) 20160
d) 64
Discussion
Explanation: The given words contains 8 letters of which E is taken 2 times.
∴ Required number of ways
$$\eqalign{ & = \frac{{8!}}{{2!}} \cr & = \frac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{2} \cr & = 20160 \cr} $$
35. In how many different ways can the letters of the word CREATE be arranged?
a) 25
b) 36
c) 360
d) 720
Discussion
Explanation: The given words contains 6 letters of which E is taken 2 times.
∴ Required number of ways
$$\eqalign{ & = \frac{{6!}}{{2!}} \cr & = \frac{{6 \times 5 \times 4 \times 3 \times 2!}}{{2!}} \cr & = 360 \cr} $$
36. In how many different ways can the letters of the word OPERATE be arranged ?
a) 360
b) 720
c) 5040
d) 2520
Discussion
Explanation: The given word contains 7 letters out of which E is taken 2 times and all other letters are different .
∴ Required number of ways
$$\eqalign{ & = \frac{{7!}}{{2!}} \cr & = \frac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{2} \cr & = 2520 \cr} $$
37. Out of 5 women and 4 men, a committee of three members is to be formed in such a way that at least one member is a women. In how many different ways can it be done ?
a) 76
b) 80
c) 84
d) 96
Discussion
Explanation: Required number of ways
$$\left( {{}^5{{\text{C}}_1} \times {}^4{{\text{C}}_2}} \right) + \left( {{}^5{{\text{C}}_2} \times {}^4{{\text{C}}_1}} \right)$$ $$ + \left( {{}^5{{\text{C}}_3}} \right)$$
$$ = \left( {5 \times \frac{{4 \times 3}}{{2 \times 1}}} \right)$$ $$ + \left( {\frac{{5 \times 4}}{{2 \times 1}} \times 4} \right)$$ $$ + \left( {\frac{{5 \times 4 \times 3}}{{3 \times 2 \times 1}}} \right)$$
$$\eqalign{ & = \left( {30 + 40 + 10} \right) \cr & = 80 \cr} $$
38. In how many ways a committee consisting of 5 men and 6 women can be formed from 8 men and 10 women ?
a) 266
b) 5040
c) 11760
d) 86400
Discussion
Explanation: Required number of ways
$$\eqalign{ & \left( {{}^8{C_5} \times {}^{10}{C_6}} \right) + \left( {{}^8{C_3} \times {}^{10}{C_4}} \right) \cr & = \frac{{8 \times 7 \times 6}}{{3!}} \times \frac{{10 \times 9 \times 8 \times 7}}{{4!}} \cr & = \frac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}} \times \frac{{10 \times 9 \times 8 \times 7}}{{4 \times 3 \times 2 \times 1}} \cr & = 11760 \cr} $$
39. In how many different ways can the letters of the word GAMBLE be arranged?
a) 15
b) 25
c) 60
d) None of these
Discussion
Explanation: The given 6 letters, all different.
∴ Required number of ways
$$\eqalign{ & {}^6{P_6} = 6! \cr & = {6 \times 5 \times 4 \times 3 \times 2 \times 1} \cr & = 720 \cr} $$
40. In how many different ways can the letters of the word RUMOUR be arranged?
a) 30
b) 90
c) 180
d) 720
Discussion
Explanation: The given word contains 6 letter out of which R is taken 2 times, U is taken to 2 times and other letters are all different.
∴ Required number of ways
$$\eqalign{ & = \frac{{6!}}{{2! \times 2!}} \cr & = \frac{{6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 2}} \cr & = 180 \cr} $$
41. From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
a) 564
b) 645
c) 735
d) 756
Discussion
Explanation: We may have (3 men and 2 women) or (4 men and 1 woman) or (5 men only).
∴ the Required number of ways
$$\eqalign{ & = \left( {{}^7{C_3} \times {}^6{C_2}} \right) + \left( {{}^7{C_4} \times {}^6{C_1}} \right) + \left( {{}^7{C_5}} \right) \cr & = \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times \frac{{6 \times 5}}{{2 \times 1}}} \right) + \left( {{}^7{C_3} \times {}^6{C_1}} \right) + \left( {{}^7{C_2}} \right) \cr & = 525 + \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times 6} \right) + \left( {\frac{{7 \times 6}}{{2 \times 1}}} \right) \cr & = \left( {525 + 210 + 21} \right) \cr & = 756 \cr} $$
42. In how many different ways can the letters of the word 'LEADING' be arranged in such a way that the vowels always come together?
a) 360
b) 480
c) 720
d) 4080
Discussion
Explanation: The word 'LEADING' has 7 different letters.
When the vowels EAI are always together, they can be supposed to form one letter.
Then, we have to arrange the letters LNDG (EAI).
Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways.
The vowels (EAI) can be arranged among themselves in 3! = 6 ways.
Therefore Required number of ways = (120 x 6) = 720
43. In how many different ways can the letters of the word 'CORPORATION' be arranged so that the vowels always come together?
a) 810
b) 1440
c) 2880
d) 50400
Discussion
Explanation: In the word 'CORPORATION', we treat the vowels OOAIO as one letter.
Thus, we have CRPRTN (OOAIO).
This has 7 (6 + 1) letters of which R occurs 2 times and the rest are different.
Number of ways arranging these letters = $$\frac{{7!}}{{2!}}$$ = 2520
Now, 5 vowels in which O occurs 3 times and the rest are different, can be arranged in $$\frac{{5!}}{{3!}}$$ = 20 ways
∴ Required number of ways = (2520 x 20) = 50400
44. Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
a) 210
b) 1050
c) 25200
d) 21400
Discussion
Explanation: Number of ways of selecting (3 consonants out of 7) and (2 vowels out of 4)
$$\eqalign{ & = \left( {{}^7{C_3} \times {}^4{C_2}} \right) \cr & = \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times \frac{{4 \times 3}}{{2 \times 1}}} \right) \cr & = 210 \cr} $$
Number of groups, each having 3 consonants and 2 vowels = 210
Each group contains 5 letters.
Number of ways of arranging 5 letters among themselves
= 5!
= 5 x 4 x 3 x 2 x 1
= 120
∴ Required number of ways = (210 x 120) = 25200
45. In how many ways can the letters of the word 'LEADER' be arranged?
a) 72
b) 144
c) 360
d) 720
Discussion
Explanation: The word 'LEADER' contains 6 letters, namely 1L, 2E, 1A, 1D and 1R.
∴ Required number of ways $$ = \frac{{6!}}{{\left( {1!} \right)\left( {2!} \right)\left( {1!} \right)\left( {1!} \right)\left( {1!} \right)}} = 360$$
46. In how many ways can the letters of the word MANAGEMENT be rearranged so that the two As do not appear together?
a) 10! - 2!
b) 9! - 2!
c) 10! - 9!
d) None of these
Discussion
Explanation: The word MANAGEMENT is a 10 letter word.
Normally, any 10 letter word can be rearranged in 10! ways.
However, as there are certain letters of the word repeating, we need to account for those. In this case, the letters A, M, E and N repeat twice each.
Therefore, the number of ways in which the letters of the word MANAGEMENT can be rearranged reduces to:
$$\frac{{10!}}{{2! \times 2! \times 2! \times 2!}}$$ The problem requires us to find out the number of outcomes in which the two As do not appear together. The number of outcomes in which the two As appear together can be found out by considering the two As as one single letter.
Therefore, there will now be only 9 letters of which three of them E, N and M repeat twice. So these 9 letters with 3 of them repeating twice can be rearranged in: $$\frac{{9!}}{{2! \times 2! \times 2!}}$$ ways. Therefore, the required answer in which the two As do not appear next to each other
47. How many numbers are there between 100 and 1000 such that at least one of their digits is 6?
a) 200
b) 225
c) 252
d) 120
Discussion
Explanation: numbers between 100 and 1000 = 900 Numbers between 100 and 1000 which do not have digit 6 in any place, = 8 × 9 × 9
= 648 Unit digit could take any value of the 9 values (0 to 9, except 6) Tens Digit could take any value of the 9 values (0 to 9, except 6) Hundreds digit could take any value of the 8 values (1 to 9, except 6) numbers between 100 and 1000 which have at least one digit as 6, = 900 - 648
= 252
48. A student is required to answer 6 out of 10 questions divided into two groups each containing 5 questions. He is not permitted to attempt more than 4 from each group. In how many ways can he make the choice?
a) 210
b) 150
c) 100
d) 200
Discussion
Explanation: Number of ways of choosing 6 from 10 = 10C6 = 210 Number of ways of attempting more than 4 from a group, = 2 × 5C5 × 5C1
= 10 Required number of ways
= 210 - 10
= 200
49. While packing for a business trip Mr. Debashis has packed 3 pairs of shoes, 4 pants, 3 half-pants,6 shirts, 3 sweater and 2 jackets. The outfit is defined as consisting of a pair of shoes, a choice of "lower wear" (either a pant or a half-pant), a choice of "upper wear" (it could be a shirt or a sweater or both) and finally he may or may not choose to wear a jacket. How many different outfits are possible ?
a) 567
b) 1821
c) 743
d) 1701
Discussion
Explanation: Number of ways a pair of shoes can be selected, = 3C1
= 3 ways Number of ways lower wear can be selected = (3 + 4) = 7 ways. Number of ways upper wear can be selected, = 3 + 6 + (3 × 6)
= 27 ways Number of ways jacket can be chosen or not chosen = 3 ways {No jacket, 1st jacket, 2nd jacket}. Total number of different outfits
= 3 × 7 × 27 × 3
= 1701 ways
50. How many positive integers 'n' can be form using the digits 3, 4, 4, 5, 6, 6, 7 if we want 'n' to exceed 60,00,000?
a) 320
b) 360
c) 540
d) 720
Discussion
Explanation: As per the given condition, number in the highest position should be either 6 or 7, which can be done in 2 ways. If the first digit is 6, the other digits can be arranged in $$\frac{{6!}}{{2!}}$$ = 360 ways. If the first digit is 7, the other digits can be arranged in $$\frac{{6!}}{{2! \times 2!}}$$ = 180 ways. Thus required possibilities for n, = 360 + 180
= 540 ways
51. A college has 10 basketball players. A 5 member's team and a captain will be selected out of these 10 players. How many different selections can be made?
a) 1260
b) 210
c) 10C6 × 6!
d) 10C5 × 6
Discussion
Explanation: A team of 6 members has to be selected from the 10 players. This can be done in 10C6 or 210 ways. Now, the captain can be selected from these 6 players in 6 ways. Therefore, total ways the selection can be made is 210 × 6 = 1260
52. How many four letter distinct initials can be formed using the alphabets of English language such that the last of the four words is always a consonant?
a) 263 × 21
b) 26 × 25 × 24 × 21
c) 25 × 24 × 23 × 21
d) None of these
Discussion
Explanation: The last of the four letter words should be a consonant. Therefore, there are 21 options. The first three letters can be either consonants or vowels. So, each of them have 26 options.
Note that the question asks you to find out the number of distinct initials and not initials where the letters are distinct. Hence, required answer
= 26 × 26 × 26 × 21
= 263 × 21
53. What is the total number of ways in which Dishu can distribute 9 distinct gifts among his 8 distinct girlfriends such that each of them gets at least one gift?
a) 72 × 8!
b) 144 × 8!
c) 36 × 8!
d) 9!
Discussion
Explanation: One among 8 gfs will get 2 gifts and remaining 7 will get one. So total of 9 gifts will be distributed among 8 gfs. i.e; 11111112 Gf who will get 2 gifts can be find out in 8C1 ways = 8 ways. Now 2 gifts can be given to selected gf in 9C2 ways. And remaining 7 gifts can be given to remaining 7 gf in 7! ways. So total no of ways= 8 × 9C2 × 7! = $$\frac{{8 \times \left( {9 \times 8} \right)}}{{2 \times 7!}}$$ = 36 × 8 × 7! = 36 × 8!
54. How many number of times will the digit 7 be written when listing the integers from 1 to 1000?
a) 270
b) 300
c) 352
d) 304
Discussion
Explanation: 7 does not occur in 1000. So we have to count the number of times it appears between 1 and 999. Any number between 1 and 999 can be expressed in the form of xyz where 0 < x, y, z < 9.
1. The numbers in which 7 occurs only once. e.g 7, 17, 78, 217, 743 etc
This means that 7 is one of the digits and the remaining two digits will be any of the other 9 digits (i.e 0 to 9 with the exception of 7)
You have 1 × 9 × 9 = 81 such numbers. However, 7 could appear as the first or the second or the third digit. Therefore, there will be 3 × 81 = 243 numbers (1-digit, 2-digits and 3- digits) in which 7 will appear only once.
In each of these numbers, 7 is written once. Therefore, 243 times.
2. The numbers in which 7 will appear twice. e.g 772 or 377 or 747 or 77
In these numbers, one of the digits is not 7 and it can be any of the 9 digits ( 0 to 9 with the exception of 7).
There will be 9 such numbers. However, this digit which is not 7 can appear in the first or second or the third place. So there are 3 × 9 = 27 such numbers.
In each of these 27 numbers, the digit 7 is written twice. Therefore, 7 is written 54 times.
3. The number in which 7 appears thrice - 777 - 1 number. 7 is written thrice in it.
Therefore, the total number of times the digit 7 is written between 1 and 999 is
= 243 + 54 + 3
= 300
55. In how many ways can 15 people be seated around two round tables with seating capacities of 7 and 8 people?
a) 15! × 8!
b) 7! × 8!
c) 15C7 × 6! × 7!
d) 2 × 15C7 × 6! × 7!
Discussion
Explanation:'n' objects can be arranged around a circle in (n - 1)! ways. If arranging these 'n' objects clockwise or counter clockwise means one and the same, then the number arrangements will be half that number. i.e., number of arrangements = $$\frac{{\left( {n - 1} \right)!}}{2}$$ You can choose the 7 people to sit in the first table in 15C7 ways. After selecting 7 people for the table that can seat 7 people, they can be seated in: (7 - 1)! = 6! The remaining 8 people can be made to sit around the second circular table in: (8 - 1)! = 7! Ways. Hence, total number of ways: 15C7 × 6! × 7!
56. A question paper consists of three sections 4,5 and 6 questions respectively. Attempting one question from each section is compulsory but a candidate need not attempt all the questions. In how many ways can a candidate attempt the questions?
a) 209
b) (4!-1) × (5!-1) × (6!-1)
c) 119
d) 29295
Discussion
Explanation: At least 1 question from each section is compulsory, so from the 1st section the candidate can attempt 1 or 2 or 3 or 4 questions. In each section each question can be dealt with in 2 ways, i.e. either he attempts it or leaves it. So far 4 question there are 2 × 2 × 2 × 2 ways to attempt.
As he has to attempt at least 1 question, the total number of ways in which he can attempt questions from 1st section is 24 - 1 Similarly for the 2nd section there are 25 - 1 ways in which he can attempt and for the 3rd section there are 26 - 1 ways. The ways in which the attempts one or more questions in any section is independent of the number of ways in which he attempts one or more questions from the other sections. Thus, total number of ways in which he can attempt questions in that paper:
= (24 - 1)(25 - 1)(26 - 1)
= 15 × 31 × 63
= 29295
57. In how many ways a President, VP and Water-boy can be selected from a group of 10 people.
a) 10C3
b) 10P3
c) 240
d) 360
Discussion
Explanation: We are selecting three different posts here, so order matters. Thus, total ways of selecting a President, VP and Water-boy from a group of 10 people would be 10P3
58. In a hockey championship, there are 153 matches played. Every two team played one match with each other. The number of teams participating in the championship is:
a) 18
b) 19
c) 17
d) 16
Discussion
Explanation: Let there were x teams participating in the games, then total number of matches, = nC2 = 153 On solving we get, n = −17 and n = 18 It cannot be negative so n = 18
59. A box contains 10 balls out of which 3 are red and rest are blue. In how many ways can a random sample of 6 balls be drawn from the bag so that at the most 2 red balls are included in the sample and no sample has all the 6 balls of the same colour?
a) 105
b) 168
c) 189
d) 120
Discussion
Explanation: Six balls can be selected in the following ways, One red ball and 5 blue balls Or Two red balls and 4 blue balls. Total number of ways, = (3C1 × 7C5) + (3C2 × 4C7)
= 63 + 105
= 168
60. Out of eight crew members three particular members can sit only on the left side. Another two particular members can sit only on the right side. Find the number of ways in which the crew can be arranged so that four men can sit on each side.
a) 864
b) 863
c) 865
d) 1728
Discussion
Explanation: Required number of ways,
= 3C2 × 4! × 4!
= 1728
61. There are five women and six men in a group. From this group a committee of 4 is to be chosen. How many different ways can a committee be formed that contain three women and one man?
a) 55
b) 60
c) 25
d) 15
Discussion
Explanation: Since, no order to the committee is mentioned, a combination instead of a permutation is used. Let's sort out what we have and what we want. Have: 5 women, 6 men. Want: 3 women AND 1 man. The word AND means multiply. Woman and Men $$\eqalign{ & ^{{\text{have}}}{{\text{C}}_{{\text{want}}}}{ \times ^{{\text{have}}}}{{\text{C}}_{{\text{want}}}} \cr & { = ^5}{{\text{C}}_3}{ \times ^6}{{\text{C}}_1} \cr & = 60 \cr} $$
62. Find the total number of distinct vehicle numbers that can be formed using two letters followed by two numbers. Letters need to be distinct.
a) 60000
b) 65000
c) 70000
d) 75000
Discussion
Explanation: Out of 26 alphabets two distinct letters can be chosen in 26P2 ways. Coming to numbers part, there are 10 ways (any number from 0 to 9 can be chosen) to choose the first digit and similarly another 10 ways to choose the second digit. Hence, there are totally 10 × 10 = 100 ways. Combined with letters there are,26P2 × 100 = 65000 ways to choose vehicle numbers.
63. In a railway compartment, there are 2 rows of seats facing each other with accommodation for 5 in each, 4 wish to sit facing forward and 3 facing towards the rear while 3 others are indifferent. In how many ways can the 10 passengers be seated?
a) 17200
b) 12600
c) 45920
d) 43200
Discussion
Explanation: The four person who wish to sit facing forward can be seated in: 5P4 ways and 3 who wish to sit facing towards the rear can be seated in: 5P3 ways and the remaining 3 can be seated in the remaining 3 seats in 3P3 ways.
Total number of ways
= 5P4 × 5P3 × 3P3
= 43200
64. There are three prizes to be distributed among five students. If no students gets more than one prize, then this can be done in :
a) 10 ways
b) 30 ways
c) 60 ways
d) 80 ways
Discussion
Explanation: 3 prize among 5 students can be distributed in,5C3 ways = 10 ways
65.A teacher of 6 students takes 2 of his students at a time to a zoo as often as he can, without taking the same pair of children together more than once. How many times does the teacher go to the zoo ?
a) 10
b) 12
c) 15
d) 20
Discussion
Explanation: Two students can be selected from 6 in 6C2 =15 ways. Therefore, the teacher goes to the zoo 15 times.
66. A man positioned at the origin of the coordinate system. the man can take steps of unit measure in the direction North, East, West or South. Find the number of ways of he can reach the point (5,6), covering the shortest possible distance.
a) 252
b) 432
c) 462
d) 504
Discussion
Explanation: In order to reach (5,6) covering the shortest distance at the same time the man has to make 5 horizontal and 6 vertical steps. The number of ways in which these steps can be taken is given by, $$\eqalign{ & = \frac{{11!}}{{5! \times 6!}} \cr & = 462 \cr} $$
67. There are 6 equally spaced points A, B, C, D, E and F marked on a circle with radius R. How many convex pentagons of distinctly different areas can be drawn using these points as vertices?
a) 6P5
b) 5
c) 3
d) None of these
Discussion
Explanation: Since, all the points are equally spaced; hence the area of all the convex pentagons will be same.
68. In an examination paper, there are two groups each containing 4 questions. A candidate is required to attempt 5 questions but not more than 3 questions from any group. In how many ways can 5 questions be selected?
a) 24
b) 48
c) 96
d) 64
Discussion
Explanation: 5 questions can be selected in the following ways, 2 question from first group and 3 question from second group Or 3 question from first group and 2 question from second group. = (4C2 × 4C3) + (3C4 × 4C2) = 24 + 24
= 48
69. After every get-together every person present shakes the hand of every other person. If there were 105 handshakes in all, how many persons were present in the party?
a) 16
b) 15
c) 13
d) 14
Discussion
Explanation: Let total number of persons present in the party be x, Then, $$\eqalign{ & \frac{{x \times \left( {x - 1} \right)}}{2} = 105 \cr & x = 15 \cr} $$
70. How many diagonals can be drawn in a pentagon?
a) 5
b) 10
c) 8
d) 7
Discussion
Explanation: A pentagon has 5 sides. We obtain the diagonals by joining the vertices in pairs. Total number of sides and diagonals, = 5C2 = $$\frac{{5 \times 4}}{{2 \times 1}}$$ = 5 × 2 = 10 This includes its 5 sides also. ⇒ Diagonals = 10 – 5 = 5 Hence, the number of diagonals
= 10 – 5
= 5
71. A family consist of a grandfather, 5 sons and daughter and 8 grandchildren. They are to be seated in a row for dinner. The grandchildren wish to occupy the 4 seats at each end and the grandfather refuses to have a grandchild on either side of him. The number of ways in which the family can be made to sit is:
a) 21530
b) 8! × 360
c) 8! × 480
d) 8! × 240
Discussion
Explanation: Total no. of seats, = 1 grandfather + 5 sons and daughters + 8 grandchildren= 14 The grandchildren can occupy the 4 seats on either side of the table in 4! = 24 ways. The grandfather can occupy a seat in (5 - 1) = 4 ways (4 gaps between 5 sons and daughter). And, the remaining seats can be occupied in 5! = 120 ways (5 seat for sons and daughter). Hence total number of required ways, = 8! × 480
72. If the letters of the word CHASM are rearranged to form 5 letter words such that none of the word repeat and the results arranged in ascending order as in a dictionary what is the rank of the word CHASM?
a) 24
b) 31
c) 32
d) 30
Discussion
Explanation: The 5 letter word can be rearranged in 5! = 120 Ways without any of the letters repeating. The first 24 of these words will start with A. Then, the 25th word will start will CA _ _ _ . The remaining 3 letters can be rearranged in 3! = 6 Ways. i.e. 6 words exist that start with CA. The next word starts with CH and then A, i.e., CHA _ _. The first of the words will be CHAMS. The next word will be CHASM. Therefore, the rank of CHASM will be 24 + 6 + 2 = 32
73. In how many ways can 5 different toys be packed in 3 identical boxes such that no box is empty, if any of the boxes may hold all of the toys ?
a) 20
b) 30
c) 25
d) 600
Discussion
Explanation: The toys are different; The boxes are identical. If none of the boxes is to remain empty, then we can pack the toys in one of the following ways: Case i. 2, 2, 1 Case ii. 3, 1, 1 Case i: Number of ways of achieving the first option 2, 2, 1 Two toys out of the 5 can be selected in 5C2 ways. Another 2 out of the remaining 3 can be selected in 3C2 ways and the last toy can be selected in 1C1 way. However, as the boxes are identical, the two different ways of selecting which box holds the first two toys and which one holds the second set of two toys will look the same. Hence, we need to divide the result by 2. Therefore, total number of ways of achieving the 2, 2, 1 option is: $$\frac{{^5{C_2}{ \times ^3}{C_2}}}{2} = \frac{{10 \times 3}}{2} = 15\,{\text{ways}}$$
Case ii: Number of ways of achieving the second option 3, 1, 1
Three toys out of the 5 can be selected in $$^5{C_3}$$ ways. As the boxes are identical, the remaining two toys can go into the two identical looking boxes in only one way.
Therefore, total number of ways of getting the 3, 1, 1 option is $$^5{C_3}$$ = 10 ways.
Total ways in which the 5 toys can be packed in 3 identical boxes
= number of ways of achieving Case i + number of ways of achieving Case ii
= 15 + 10
= 25 ways.
74.What is the value of 1 × 1! + 2 × 2! + 3 × 3! + . . . . . . . . n × n!
where n! means n factorial or n(n-1) (n-2) . . . . . . . . 1
a) n × (n - 1) × (n - 1)!
b) (n + 1)! - {n × (n - 1)}
c) (n + 1)! - n!
d) (n + 1)! - 1!
Discussion
Explanation: 1 × 1! = (2 - 1) × 81! = 2 × 1! - 1 × 1! = 2! - 1!
2 × 2! = (3 - 1) × 2! = 3 × 2! - 2! = 3! - 2!
3 × 3! = (4 - 1) × 3! = 4 × 3! - 3! = 4! - 3!
..
..
..
n × n! = (n + 1 - 1) × n! = (n + 1) (n!) - n! = (n + 1)! - n!
Summing up all these terms, we get (n + 1)! - 1!
75. When six fair coins are tossed simultaneously, in how many of the outcomes will at most three of the coins turn up as heads?
a) 25
b) 41
c) 22
d) 42
Discussion
Explanation: The question requires you to find number of the outcomes in which at most 3 coins turn up as heads. i.e., 0 coins turn heads or 1 coin turns head or 2 coins turn heads or 3 coins turn heads. The number of outcomes in which 0 coins turn heads is, 6C0 = 1 outcome. The number of outcomes in which 1 coin turns head is, 6C1 = 6 outcomes. The number of outcomes in which 2 coins turn heads is, 6C2 = 15 outcomes. The number of outcomes in which 3 coins turn heads is, 6C3 = 20 outcomes. Therefore, total number of outcomes
= 1 + 6 + 15 + 20
= 42 outcomes.
76. In how many ways can 8 Indians and, 4 American and 4 Englishmen can be seated in a row so that all person of the same nationality sit together?
a) 3! 4! 8! 4!
b) 3! 8!
c) 4! 4!
d) 8! 4! 4!
Discussion
Explanation: Taking all person of same nationality as one person, then we will have only three people.
These three person can be arranged themselves in 3! Ways.
8 Indians can be arranged themselves in 8! Way.
4 American can be arranged themselves in 4! Ways.
4 Englishman can be arranged themselves in 4! Ways.
Hence, required number of ways = 3! 8! 4! 4! Ways.
77. How many Permutations of the letters of the word APPLE are there?
a) 600
b) 120
c) 240
d) 60
Discussion
Explanation: APPLE = 5 letters. But two letters PP is of same kind. Thus, required permutations, $$\eqalign{ & = \frac{{5!}}{{2!}} \cr & = \frac{{120}}{2} \cr & = 60 \cr} $$
78. How many different words can be formed using all the letters of the word ALLAHABAD ?
(a) When vowels occupy the even positions.
(b) Both L do not occur together.
a) 7560,60,1680
b) 7890,120,650
c) 7650,200,4444
d) None of these
Discussion
Explanation: ALLAHABAD = 9 letters. Out of these 9 letters there is 4 A's and 2 L's are there. So, permutations = $$\frac{{9!}}{{4!.2!}}$$ = 7560 (a) There are 4 vowels and all are alike i.e. 4A's. _2nd _4th _6th _8th _ These even places can be occupied by 4 vowels. In $$\frac{{4!}}{{4!}}$$ = 1 Way. In other five places 5 other letter can be occupied of which two are alike i.e. 2L's. Number of ways = $$\frac{{5!}}{{2!}}$$ Ways. Hence, total number of ways in which vowels occupy the even places = $$\frac{{5!}}{{2!}}$$ × 1 = 60 ways. (b) Taking both L's together and treating them as one letter we have 8 letters out of which A repeats 4 times and others are distinct. These 8 letters can be arranged in $$\frac{{8!}}{{4!}}$$ = 1680 ways. Also two L can be arranged themselves in 2! ways. So, Total no. of ways in which L are together = 1680 × 2 = 3360 ways. Now, Total arrangement in which L never occur together, = Total arrangement - Total no. of ways in which L occur together. = 7560 - 3360
= 4200 ways
79. In how many ways can 10 examination papers be arranged so that the best and the worst papers never come together?
a) 8 × 9!
b) 8 × 8!
c) 7 × 9!
d) 9 × 8!
Discussion
Explanation: No. of ways in which 10 paper can arranged is 10! Ways. When the best and the worst papers come together, regarding the two as one paper, we have only 9 papers. These 9 papers can be arranged in 9! Ways. And two papers can be arranged themselves in 2! Ways. No. of arrangement when best and worst paper do not come together, = 10! - 9! × 2!
= 9!(10 - 2)
= 8 × 9!
80. In how many ways 4 boys and 3 girls can be seated in a row so that they are alternate.
a) 144
b) 288
c) 12
d) 256
Discussion
Explanation: Let the Arrangement be,
B G B G B G B
4 boys can be seated in 4! Ways
Girl can be seated in 3! Ways
Required number of ways,
= 4! × 3!
= 144
81. How many arrangements of four 0's (zeroes), two 1's and two 2's are there in which the first 1 occur before the first 2?
a) 420
b) 360
c) 320
d) 210
Discussion
Explanation: Total number of arrangements = $$\frac{{8!}}{{4! \times 2! \times 2!}}$$ = 420 Since, there are two 1's and two 0's, the number of arrangements in which the first 1 is before the first 2 is same as the number of arrangement in which the first 2 is before the first 1 and they are each equal to half the total number of arrangements = 210
82. How many different five-letter words can be formed using the letter from the world APPLE?
a) 24
b) 60
c) 120
d) 240
Discussion
Explanation: If the two P’s were distinct (they could have different subscripts and colours), the number of possible permutations would have been 5! = 120 For example let us consider one permutation: P1LEAP2 Now if we permute the P’s amongst them we still get the same word PLEAP. The two P’s can be permuted amongst them in 2! ways. We were counting P1LEAP2 and P2LEAP1 as different arrangements only because we were artificially distinguishing between the two P’s Hence the number of different five letter words that can be formed is: = $$\frac{{5!}}{{\left(1!\right) \left(2!\right) \left(1!\right) \left(1!\right)}}$$
= 60
83.I have an amount of Rs. 10 lakh, which I went to invest in stocks of some companies. I always invest only amounts that are multiples of Rs 1 lakh in the stock of any company. If I can choose from among the stocks of five different companies, In how many ways can I invest the entire amount that I have?
a) 252
b) 250
c) 1001
d) 1089
Discussion
Explanation: The situation is similar to placing 10 identical balls among 5 distinguishable boxes, where a box may have zero or more balls in it. This case can be represented as arranging ten balls and (5 - 1) four walls in the single row, which can be done in 14C4 ways. (The balls placed between every successive pair of walls belong to one group) 14C4 = 1001 ways.
84. A selection committee is to be chosen consisting of 5 ex-technicians. Now there are 12 representatives from four zones. It has further been decided that if Mr. X is selected, Y and Z will not be selected and vice-versa. In how many ways it can be done?
a) 572
b) 672
c) 472
d) 372
Discussion
Explanation: 10C5 : when both are not included. 10C4 : when one of them is included. Number of ways
= 10C5 + 10C4 + 10C4
= 672
85. How many 5-digit positive integers exist the sum of whose digits are odd?
a) 36000
b) 38000
c) 45000
d) 90000
Discussion
Explanation: There are 9 × 104 = 90000, 5-digit positive integers. Out of these 90000 positive integers, the sum of the digits of half of the numbers will add up to an odd number and the remaining half will add up to an even number. Hence, there are $$\frac{{90000}}{2}$$ = 45000, 5-digit positive integers whose sum add up to an odd number.
86. A two member committee comprising of one male and one female member is to be constitute out of five males and three females. Amongst the females. Ms. A refuses to be a member of the committee in which Mr. B is taken as the member. In how many different ways can the committee be constituted
a) 11
b) 12
c) 13
d) 14
Discussion
Explanation: 5C1 × 3C1 - 1 = 15 - 1 = 14
87. In how many ways 2 students can be chosen from the class of 20 students?
a) 190
b) 180
c) 240
d) 360
Discussion
Explanation: Number of ways
$$\eqalign{ & { = ^{20}}{{\text{C}}_2} \cr & = \frac{{20!}}{{2! \times 18!}} \cr & = 20 \times \frac{{19}}{2} \cr & = 190 \cr} $$
88. Three gentlemen and three ladies are candidates for two vacancies. A voter has to vote for two candidates. In how many ways can one cast his vote?
a) 9
b) 30
c) 36
d) 15
Discussion
Explanation: There are 6 candidates and a voter has to vote for any two of them. So, the required number of ways is, $$\eqalign{ & { = ^6}{{\text{C}}_2} \cr & = \frac{{6!}}{{2! \times 4!}} \cr & = 15 \cr} $$
89. A question paper has two parts, A and B, each containing 10 questions. If a student has to choose 8 from part A and 5 from part B, in how many ways can he choose the questions?
a) 11340
b) 12750
c) 40
d) 320
Discussion
Explanation: There 10 questions in part A out of which 8 question can be chosen as = 10C8 Similarly, 5 questions can be chosen from 10 questions of Part B as = 10C5 Hence, total number of ways, $$\eqalign{ & { = ^{10}}{{\text{C}}_8}{ \times ^{10}}{{\text{C}}_5} \cr & = \frac{{10!}}{{2! \times 8!}} \times \frac{{10!}}{{5! \times 5}} \cr & = \left\{ {10 \times \frac{9}{2}} \right\} \times \left\{ {\frac{{10 \times 9 \times 8 \times 7 \times 6}}{{5 \times 4 \times 3 \times 2 \times 1}}} \right\} \cr & = 11340 \cr} $$
90. Find the number of triangles which can be formed by joining the angular points of a polygon of 8 sides as vertices.
a) 56
b) 24
c) 16
d) 8
Discussion
Explanation: A triangle needs 3 points. And polygon of 8 sides has 8 angular points. Hence, number of triangle formed, $$\eqalign{ & { = ^8}{{\text{C}}_3} \cr & = \frac{{8 \times 7 \times 6}}{{1 \times 2 \times 3}} \cr & = 56 \cr} $$
91. The letter of the word LABOUR are permuted in all possible ways and the words thus formed are arranged as in a dictionary. What is the rank of the word LABOUR?
a) 242
b) 240
c) 251
d) 275
Discussion
Explanation: The order of each letter in the dictionary is ABLORU. Now, with A in the beginning, the remaining letters can be permuted in 5! ways. Similarly, with B in the beginning, the remaining letters can be permuted in 5! ways. With L in the beginning, the first word will be LABORU, the second will be LABOUR. Hence, the rank of the word LABOUR is, 5! + 5! + 2 = 120 + 120 + 2 = 242 Note: 5! = 5 × 4 × 3 × 2 × 1 = 120
92. A class photograph has to be taken. The front row consists of 6 girls who are sitting. 20 boys are standing behind. The two corner positions are reserved for the 2 tallest boys. In how many ways can the students be arranged?
a) 6! × 1440
b) 18! × 1440
c) 18! × 2! × 1440
d) None of these
Discussion
Explanation: Two tallest boys can be arranged in 2! ways, rest 18 can be arranged in 18! ways. Girls can be arranged in 6! ways. Total number of ways in which all the students can be arranged, = 2! × 18! × 6! = 18! × 1440 Note: N! = N × (N - 1) × (N - 2) × . . . . × 1 So, 18! = 18 × 17 × 16 × 15 . . . . . . . . × 1
93. If $$5{ \times ^{\text{n}}}{{\text{P}}_3} = 4{ \times ^{\left( {{\text{n}} + 1} \right)}}{{\text{P}}_{3,}}$$ find n?
a) 10
b) 11
c) 12
d) 14
Discussion
Explanation: nP3 = n × (n–1) × (n–2) (n+1)P3 = (n+1) × n × (n–1) Now, 5 × n × (n–1) × (n–2) = 4 × (n+1) × n × (n–1) Or, 5(n−2) = 4(n+1) Or, 5n − 10 = 4n + 4 Or, 5n − 4n = 4 + 10 Hence, n = 14 Note: nPr = $$\frac{{{\text{n}}!}}{{\left( {{\text{n}} - {\text{r}}} \right)!}}$$
94. Ten different letters of alphabet are given, words with 5 letters are formed from these given letters. Then, the number of words which have at least one letter repeated is:
a) 69760
b) 30240
c) 99748
d) 42386
Discussion
Explanation: Number of words which have at least one letter replaced, = Total number of words - total number of words in which no letter is repeated. = 105 – 16P5 = 100000 − 30240
= 69760
95. 12 chairs are arranged in a row and are numbered 1 to 12. 4 men have to be seated in these chairs so that the chairs numbered 1 to 8 should be occupied and no two men occupy adjacent chairs. Find the number of ways the task can be done.
a) 360
b) 384
c) 432
d) 470
Discussion
Explanation: Given there are 12 numbered chairs, such that chairs numbered 1 to 8 should be occupied. 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12 The various combinations of chairs that ensure that no two men are sitting together are listed. (1, 3, 5, __ ), The fourth chair can be 5,6,10,11 or 12, hence 5 ways. (1, 4, 8, __ ), The fourth chair can be 6,10,11 or 12 hence 4 ways. (1, 5, 8, __ ), the fourth chair can be 10,11 or 12 hence 3 ways. (1, 6, 8, __ ), the fourth chair can be 10,11 or 12 hence 3 ways. (1, 8, 10, 12) is also one of the combinations. Hence, 16 such combinations exist. In case of each these combinations we can make the four men inter arrange in 4! ways. Hence, the required result =16 × 4! = 384
96. In how many ways can the letters of the word EDUCATION be rearranged so that the relative position of the vowels and consonants remain the same as in the word EDUCATION?
a) 9! × 4
b) 9! × 4! × 5!
c) 4! × 5!
d) None of these
Discussion
Explanation: The word EDUCATION is a 9 letter word, with none of the letters repeating. The vowels occupy 3rd, 5th, 7th and 8th position in the word and the remaining 5 positions are occupied by consonants. As the relative position of the vowels and consonants in any arrangement should remain the same as in the word EDUCATION, the vowels can occupy only the before mentioned 4 places and the consonants can occupy 1st, 2nd, 4th, 6th and 9th positions. The 4 vowels can be arranged in the 3rd, 5th, 7th and 8th position in 4! Ways. Similarly, the 5 consonants can be arranged in 1st, 2nd, 4th, 6th and 9th
97.A team of 8 students goes on an excursion, in two cars, of which one can seat 5 and the other only 4. In how many ways can they travel?
a) 392
b) 126
c) 26
d) 9
Discussion
Explanation: There are 8 students and the maximum capacity of the cars together is 9. We may divide the 8 students as follows:
Case I: 5 students in the first car and 3 in the second. Hence, 8 students are divided into groups of 5 and 3 in 8C3 = 56 ways.
Case II: 4 students in the first car and 4 in the second. So, 8 students are divided into two groups of 4 and 4 in 8C4 = 78 ways. Therefore, the total number of ways in which 8 students can travel is: 56 + 70 = 126
98. A tea expert claims that he can easily find out whether milk or tea leaves were added first to water just by tasting the cup of tea. In order to check this claims 10 cups of tea are prepared, 5 in one way and 5 in other. Find the different possible ways of presenting these 10 cups to the expert.
a) 252
b) 240
c) 300
d) 340
Discussion
Explanation: Since, there are 5 cups of each kind, prepared with milk or tea leaves added first, are identical hence, total number of different people ways of presenting the cups to the expert is, $$\eqalign{ & = \frac{{10!}}{{5! \times 5!}} \cr & = 252 \cr} $$
99. A committee is to be formed comprising 7 members such that there is a simple majority of men and at least 1 woman. The shortlist consists of 9 men and 6 women. In how many ways can this committee be formed?
a) 4914
b) 3630
c) 3724
d) 3824
Discussion
Explanation: Three possibilities: (1W+6M), (2W+5M), (3W+4M) = (6C1 × 9C6) + (6C2 × 9C5) + (6C3 × 9C4)
= 4914
100. How many alphabets need to be there in a language if one were to make 1 million distinct 3 digit initials using the alphabets of the language ?
a) 100
b) 50
c) 26
d) 1000
Discussion
Explanation: 1 million distinct 3 digit initials are needed. Let the number of required alphabets in the language be 'n'. Therefore, using 'n' alphabets we can form n × n × n = n3 distinct 3 digit initials.
NOTE: Distinct initials are different from initials where the digits are different. For instance, AAA and BBB are acceptable combinations in the case of distinct initials while they are not permitted when the digits of the initials need to be different. This n3 different initials = 1 million. i.e. n3 = 106 (1 million =106) n3 = 1023 n = 102
n = 100 Hence, the language needs to have a minimum of 100 alphabets