Question

The number of ways that a committee consisting of $5$ men and $3$ women can be formed from a group of $7$ men and $6$ women so that $2$ particular women both never be in the committee is

• A. $420$
• B. $5040$
• C. $336$
• D. $216$

Answer 1

The correct option is C $336$

Number of ways${=}^7{C}_5{\times }^6{C}_3{-}^4{C}_1{\times }^2{C}_2$

${=}^7{C}_5\left({}^6{C}_3-4\times 1\right)$

$=\frac{7\times 6}{2}\left(\frac{6\times 5\times 4}{3\times 2}-4\right)$

$=21\left(20-4\right)$

$=21\times 16$

$=336$