a) 564
b) 645
c) 735
d) 756
Answer: d
Explanation: Required number of ways
$$ = \left( {{}^7{C_3} \times {}^6{C_2}} \right) + $$ $$\left( {{}^7{C_4} \times {}^6{C_1}} \right) + $$ $$\left( {{}^7{C_5} \times {}^6{C_0}} \right)$$
$$ = \left\{ {\frac{{7 \times 6 \times 5}}{{3!}} \times \frac{{6 \times 5}}{{2!}}} \right\}$$ $$ + \left( {{}^7{C_3} \times {}^6{C_1}} \right)$$ $$ + \left( {{}^7{C_2} \times 1} \right)$$
$$ = \left\{ {\frac{{7 \times 6 \times 5}}{6} \times \frac{{6 \times 5}}{{2 \times 1}}} \right\}$$ $$ + \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times 6} \right)$$ $$ + \left( {\frac{{7 \times 6}}{{2 \times 1}} \times 1} \right)$$
$$\eqalign{
& = \left( {525 + 210 + 21} \right) \cr
& = 756 \cr} $$
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