Answer: Option C
The committee should have
(1 man, 4 women) or (2 men, 3 women) or (3 men, 2 women) or ( 4 men, 1 woman)
Required number of ways
$$ = \left( {{}^4{C_1} \times {}^5{C_4}} \right) + \left( {{}^4{C_2} \times {}^5{C_3}} \right)$$      $$ + \left( {{}^4{C_3} \times {}^5{C_2}} \right)$$   $$ + \left( {{}^4{C_6} \times {}^5{C_1}} \right)$$
$$ = \left( {{}^4{C_1} \times {}^5{C_1}} \right) + \left( {{}^4{C_2} \times {}^5{C_2}} \right)$$      $$ + \left( {{}^4{C_1} \times {}^5{C_2}} \right)$$   $$ + \left( {{}^4{C_4} \times {}^5{C_1}} \right)$$
$$ = \left( {4 \times 5} \right) + \left( {\frac{{4 \times 3}}{{2 \times 1}} \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$      $$ + \left( {4 \times \frac{{5 \times 4}}{{2 \times 1}}} \right)$$   $$ + \left( {1 \times 5} \right)$$
$$ = \left( {20 + 60 + 40 + 5} \right)$$
$$ = 125$$