In how many ways can a committee be formed of $5$ members from $6$ men and $4$ women if the committee has at least one woman

The correct option is D. $246$

Given: number of men $=6$

Number of women $=4$

We have to form a committee be formed of $5$ members.

The number of ways can be deduce as follows :
$1$ woman and $4$ men ${=}^4{C}_1{\times }^6{C}_4=4{\times }^6{C}_2=4\times \frac{6\times 5}{1\times 2}=4\times 15=60$
$2$ women and $3$ men ${=}^4{C}_2{\times }^6{C}_3=\frac{4\times 3}{1\times 2}\times \frac{6\times 5\times 4}{1\times 2\times 3}=6\times 20=120$
$3$ women and $2$ men ${=}^4{C}_3{\times }^6{C}_2=\frac{4\times 3\times 2}{1\times 2\times 3}\times \frac{6\times 5}{1\times 2}=4\times 15=60$
$4$ women and $1$ man ${=}^4{C}_4{\times }^6{C}_1=1\times 6=6$
Required number of ways $60+120+60+6=246.$