Question

There are $6$ batsmen, $7$ bowlers, and $2$ wickets keepers. Find the number of ways of forming a team of $11$ players having at least $4$ batsmen, at least $5$ bowlers, and at least $1$ wicket keeper.

• A. $567$
• B. $525$
• C. $462$
• D. $777$

Answer 1

The correct option is D

$777$

Find the number of ways

Given, team size $=11$ players

Case $1$: $5$ batsmen, $5$ bowlers, $1$ wicket-keepers.

Number of ways

$$\begin{gathered} ={}^6{\mathrm{C}}_5\times {}^7{\mathrm{C}}_5\times {}^2{\mathrm{C}}_1 \\ =6\times \frac{7!}{5!\left(7-5\right)!}\times 2...\left({}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}\right) \\ =6\times \frac{7\times 6\times 5!}{5!\times 2}\times 2 \\ =252 \end{gathered}$$

Case $2$: $4$ batsmen, $6$ bowlers, $1$ wicket-keepers.

Number of ways

$$\begin{gathered} ={}^6{\mathrm{C}}_4\times {}^7{\mathrm{C}}_6\times {}^2{\mathrm{C}}_1 \\ =\frac{6!}{4!\left(6-4\right)!}\times 7\times 2...\left({}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}\right) \\ =\frac{6\times 5\times 4!}{4!\times 2}\times 7\times 2 \\ =210 \end{gathered}$$

Case $3$: $4$ batsmen, $5$ bowlers, $2$ wicket-keepers.

Number of ways

$$\begin{gathered} ={}^6{\mathrm{C}}_4\times {}^7{\mathrm{C}}_5\times {}^2{\mathrm{C}}_2 \\ =\frac{6!}{4!\left(6-4\right)!}\times \frac{7!}{5!\left(7-5\right)!}\times 1...\left({}^n{C}_r=\frac{n!}{r!\left(n-r\right)!},{}^n{C}_n=1\right) \\ =\frac{6\times 5\times 4!}{4!\times 2}\times \frac{7\times 6\times 5!}{5!\times 2}\times 1 \\ =315 \end{gathered}$$

Total number of ways $=252+210+315=777$

Hence option (D), $777$ is the correct answer.