Question

The number of ways in which a committee consisting of 6 men and 3 women may be formed from a group of 10 men and 6 women is:

• A. ${}^{16}{C}_9$
• B. ${}^{10}{C}_6$
• C. ${}^6{C}_3$
• D. ${}^{10}{C}_4{.}^6{C}_3$

Answer 1

Answer: (D)

${}^{10}{C}_4{.}^6{C}_3$

6 men from 10 can be selected in ${}^{10}{C}_6$ ways
And 3 women from 6 can be selected in ${}^6{C}_3$ ways
Hence by multiplication rule,
The required number of ways ${=}^{10}{C}_6$ ${\times }^6{C}_3$

Hence option D is correct ($sinc{e}^{10}{C}_6{=}^{10}{C}_4$)