Combination questions Q: A team of $11$ is to be chosen out of $15$ cricketers of whom $5$ are bowlers and $2$ others are wicket keepers. In how many ways can this be done so that the team contains at least $4$ bowlers and at least $1$ wicket keeper?

The number of ways to choose exactly $4$ bowlers and exactly $1$ keeper:
$$\binom{5}{4}\cdot\binom{2}{1}\cdot\binom{15-5-2}{11-4-1}=280$$
The number of ways to choose exactly $5$ bowlers and exactly $1$ keeper:
$$\binom{5}{5}\cdot\binom{2}{1}\cdot\binom{15-5-2}{11-5-1}=112$$
The number of ways to choose exactly $4$ bowlers and exactly $2$ keepers:
$$\binom{5}{4}\cdot\binom{2}{2}\cdot\binom{15-5-2}{11-4-2}=280$$
The number of ways to choose exactly $5$ bowlers and exactly $2$ keepers:
$$\binom{5}{5}\cdot\binom{2}{2}\cdot\binom{15-5-2}{11-5-2}=70$$
The number of ways to choose at least $4$ bowlers and at least $1$ keeper:
$$280+112+280+70=742$$
