Question

A cricket $11$ is to be chosen from $16$ players of whom $7$ are bowlers, $4$ can do the wicketkeeping. Number of ways this can be done to contain exactly $5$ bowlers, $2$ wicket keepers is

• A. $945$
• B. $885$
• C. $630$
• D. $715$

Answer 1

Answer: (C)

$630$

In $16$ players $-7$ are bowlers, $4$ wicket keepers

so $5$ are batsmen $(16-7+4)$

So we want $11$ players - In which $5$ between $2$ wicket keepers,

so $(11-5+2)=2$ are batsmen (non bowlers & w.k)

$\Rightarrow {}^7{C}_5{\times }^4{C}_2{\times }^5{C}_4=\frac{7!}{5!(7-5)!}\times \frac{4!}{2!(4-2)!}\times \frac{5!}{4!(5-4)!}$

$=\frac{6\times 7}{2}\times \frac{3\times 4}{2}\times \frac{5}{1}$

$=21\times 6\times 5=21\times 30$

$630$

So In $630$ ways we can be done that contains exactly $3$ bowlers, $2$ wicket keepers.