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 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?
Here, order does not matter. Therefore, we need to count combinations. 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 $=\,^{5} C _{3}=\frac{5 !}{3 ! 2 !}=\frac{4 \times 5}{2}=10.$
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
$=\,^{2} C_{1} \times^{3} C_{2}=\frac{2 !}{1 ! 1 !} \times \frac{3 !}{2 ! 1 !}=6$
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