Question

The number of ways in which $10$ students can be divided into three teams, one containing $4$ and others $3$ each, is

• A. $\frac{10!}{4!3!3!}$
• B. $2100$
• C. ${}^{10}{C}_4{.}^5{C}_2$
• D. $\frac{10!}{6!3!3!}.\frac{1}{2}$

Answer 1

Answer: (B), (C)

$2100$
${}^{10}{C}_4{.}^5{C}_2$

The total number of ways to form three teams is same as selecting $4$ students out of $10$ to form first team, then $3$ students from remaining $6$ to form second team and the remaining $3$ forms another team.

This can be done as follows

$=3\times \left(\frac{{}^{10}{C}_4{}^6{C}_3}{3!}\right)=\frac{10!6!}{4!6!3!3!3!}\times 3={}^{10}{C}_4{}^5{C}_2=2100$