Question

There are $5$ girls and $7$ boys. A team of $3$ boys and $2$ girls is to be formed such that no two specific boys are in the same team. Number of way to do so are

• A. $400$
• B. $250$
• C. $200$
• D. $300$

Answer 1

The correct option is D

$300$

Find the number of ways for the given conditions

Given, $5$ girls and $7$ boys

Total numbers of team $={}^7{\mathrm{C}}_3\times {}^5{\mathrm{C}}_2$

$$\begin{gathered} =\frac{7!}{3!\left(7-3\right)!}\times \frac{5!}{2!\left(5-2\right)!}\left({}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}\right) \\ =\frac{7!}{3!\times 4!}\times \frac{5!}{2!\times 3!} \\ =\frac{7\times 6\times 5}{6}\times \frac{5\times 4}{2} \\ =350 \end{gathered}$$

Number of teams with those two specific boys $={}^5{\mathrm{C}}_1\times {}^5\mathrm{C}_2$

$$\begin{gathered} =5\times \frac{5!}{2!\left(5-2\right)!}\left({}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!\left(\mathrm{n}-\mathrm{r}\right)!}\right) \\ =5\times \frac{5\times 4\times 3!}{2\times 3!} \\ =50 \end{gathered}$$

Required number of ways $=350-50=300$

Hence, option D is correct .