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**Hint:**In this question we are given a total 16 players out of which 4 are bowlers and 2 are wicket keepers. We need to find the number of ways to form a team of 11 players having 3 bowlers and 1 wicket keeper. For this, we will first find non-bowlers, non-wicket keeper players. Then we will find ways of choosing 3 bowlers out of 4, then 1 wicket keeper out of 2 and at last 7 non-bowler, non-wicket keeper players out of the remaining. We will use a combination method here. Number of ways of selecting r items out of n is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.

**Complete step-by-step solution:**

Here we are given the total players as 16. Number of bowlers are 4 and number of wicket keepers are 2. Remaining players (non-bowler, non-wicket keeper) will be 16 - (4+2) = 16-6 = 10.

Now we need to form a team of 11 players with 3 bowlers and 1 wicket keeper. So let us find ways of forming such a team.

We know that the total number of ways of selecting r items out of n items is given by ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.

So number of ways of choosing 3 bowlers out of 4 bowlers will be given by ${}^{4}{{C}_{3}}=\dfrac{4!}{3!\left( 4-3 \right)!}=\dfrac{4\times 3!}{3!\times 1!}$.

Number of ways of choosing 3 bowlers = 4.

Now let us find number of ways of choosing 1 wicket keeper out of 2 wicket keepers we get, ${}^{2}{{C}_{1}}=\dfrac{2!}{1!1!}=2$.

Number of ways of choosing 1 wicket keeper = 1.

We are left with 10 players and we have to choose 11 - (3+1) = 7 players out of them.

So number of ways of choosing 7 non-bowler non-wicket keeper out of 10 are:

${}^{10}{{C}_{7}}=\dfrac{10!}{7!\left( 10-7 \right)!}=\dfrac{10!}{7!3!}=\dfrac{10\times 9\times 8\times 7!}{7!\times 3\times 2}=120$.

Therefore total number of ways of selecting a team of 11 players = ways of selecting bowlers × ways of selecting wicket keeper × ways of selecting other players.

$4\times 2\times 120=960$.

**Hence the total number of ways of selecting a team becomes 960.**

**Note:**Students should consider every possibility before giving a final answer. Note that we have used multiplication at the end because all these events are occurring simultaneously. Take care while solving ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.

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