A select group of 4 is to be formed from 8 men and 6 women in such a way that the group must have at least 1 women. In how many different ways can it be done ?

Required number of ways
$$ = \left( {{}^6{C_1} \times {}^8{C_3}} \right) + \left( {{}^6{C_2} \times {}^8{C_2}} \right)$$     $$ + \left( {{}^6{C_3} \times {}^8{C_1}} \right)$$   $$ + \left( {{}^6{C_4} \times {}^8{C_0}} \right)$$
$$ = \left\{ {6 \times \frac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}}} \right\} + $$    $$\left( {\frac{{6 \times 5}}{{2 \times 1}} \times \frac{{8 \times 7}}{{2 \times 1}}} \right)$$   $$ + \left( {\frac{{6 \times 5 \times 4}}{{3 \times 2 \times 1}} \times 8} \right)$$    $$ + \left( {{}^6{C_2} \times 1} \right)$$
$$ = \left\{ {6 \times \frac{{8 \times 7 \times 6}}{{3 \times 2 \times 1}}} \right\}$$    $$ +\, 420\, + $$  $$\left( {\frac{{6 \times 5 \times 4}}{6} \times 8} \right)$$   $$ + \left( {\frac{{6 \times 5}}{{2 \times 1}} \times 1} \right)$$
$$ = \left( {336 + 420 + 160 + 15} \right)$$
$$ = 931$$