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?

Required number of ways
$$ = \left( {{}^7{C_3} \times {}^6{C_2}} \right) + $$   $$\left( {{}^7{C_4} \times {}^6{C_1}} \right) + $$   $$\left( {{}^7{C_5} \times {}^6{C_0}} \right)$$
$$ = \left\{ {\frac{{7 \times 6 \times 5}}{{3!}} \times \frac{{6 \times 5}}{{2!}}} \right\}$$     $$ + \left( {{}^7{C_3} \times {}^6{C_1}} \right)$$   $$ + \left( {{}^7{C_2} \times 1} \right)$$
$$ = \left\{ {\frac{{7 \times 6 \times 5}}{6} \times \frac{{6 \times 5}}{{2 \times 1}}} \right\}$$     $$ + \left( {\frac{{7 \times 6 \times 5}}{{3 \times 2 \times 1}} \times 6} \right)$$    $$ + \left( {\frac{{7 \times 6}}{{2 \times 1}} \times 1} \right)$$
$$\eqalign{ & = \left( {525 + 210 + 21} \right) \cr & = 756 \cr} $$