Question

Each of the $3$ committees has $1$ vacancy can be filled from a group of $6$ people. Find the number of ways the $3$ vacancies can be filled if each person can serve atmost one committee.

Answer 1

Select one committee out of $3{=}^3{C}_1$

Select one person out of $6{=}^6{C}_1$

Select one committee out of $2{=}^2{C}_1$

Select one person out of $5{=}^5{C}_1$

Select last committee$=1$

Select one person out of $4{=}^4{C}_1$

number of ways the $3$ vacancies can be filled if Each person can serve on atmost $1$ committee

${=}^3{C}_1{\times }^6{C}_1{\times }^2{C}_1{\times }^5{C}_1\times 1{\times }^4{C}_1$

$=\frac{3!}{2!1}\times \frac{6!}{5!1!}\times \frac{2!}{1}\times \frac{5!}{4!}\times \frac{4!}{3!1}$

$=6!=720$ways.