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
Answer: d
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
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