Question

A committee of $3$ persons is to be constituted from a group of $2$ men and $3$ women. In how many ways can this be done? How many of these committees would consist of $1$ man and $2$ women?

• A. $10, 4$
• B. $10, 6$
• C. $5, 4$
• D. $6, 4$

Answer 1

Answer: (B)

There will be as many committees as there are combinations of $5$ different persons taken $3$ at a time. Hence, the required number of ways
$=\,{}^{5}C_{3} = \frac{5!}{3! \,2! }$
$= \frac{4\times5}{2} = 10$.
Now, $1$ man can be selected from $2$ men in $^{2}C_{1}$ ways and $2$ women can be selected from $3$ women in $^{3}C_{2}$ ways. Therefore, the required number of committees
$= \,{}^{2}C_{1} \times\,{}^{3}C_{2} $
$= \frac{2!}{1! \,1!} \times \frac{3!}{2! \,1!} = 6$