Answer

Verified

387.3k+ views

**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 ${}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$.

Recently Updated Pages

Cryolite and fluorspar are mixed with Al2O3 during class 11 chemistry CBSE

Select the smallest atom A F B Cl C Br D I class 11 chemistry CBSE

The best reagent to convert pent 3 en 2 ol and pent class 11 chemistry CBSE

Reverse process of sublimation is aFusion bCondensation class 11 chemistry CBSE

The best and latest technique for isolation purification class 11 chemistry CBSE

Hydrochloric acid is a Strong acid b Weak acid c Strong class 11 chemistry CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Define limiting molar conductivity Why does the conductivity class 12 chemistry CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Name 10 Living and Non living things class 9 biology CBSE

The Buddhist universities of Nalanda and Vikramshila class 7 social science CBSE

Write a letter to the principal requesting him to grant class 10 english CBSE