Question

In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?

Answer 1

Hint- In this question, a cricket team of 11 players is to be selected from 17 players. Only 5 players in the 17 players can bowl so the remaining 12 players are batsman. To form the team of 11 players in which 4 bowlers should be there, we will find the number of ways to select 4 bowlers from 5 bowlers and remaining 7 players from the 12 batsman and use the fundamental principle of counting or multiplication rule after that to get the answer.

Complete step-by-step answer:
It is given that we have to select 11 players from the 17 players.
Also, it is given that the 17 players consist of 5 bowlers. So, the remaining 12 players are batsmen.
Now, the 11-player team must consist of 4 bowlers. So, we will find a number of ways to select 4 bowlers from the 5 bowlers and the remaining 7 players from the 12 batsmen.
4 bowlers can be selected from 5 bowlers in ${}^5{C_4}$ ways and the remaining 7 players can be selected from remaining 12 batsmen in ${}^{12}{C_7}$ ways.
Now, we know that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ where r is the number of unordered outcomes and n is the number of possibilities to choose from.
By Fundamental principle of counting or the Multiplication or product rule, required number of ways to select the cricket team,
$\Rightarrow {}^5{C_4} \times {}^{12}{C_7} = \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}} \times \dfrac{{12!}}{{7!\left( {12 - 7} \right)!}} = \dfrac{{5!}}{{4!1!}} \times \dfrac{{12!}}{{7!5!}} = \dfrac{{5 \times 4!}}{{4!}} \times \dfrac{{12 \times 11 \times 10 \times 9 \times 8 \times 7!}}{{7!5!}}$
$= 5 \times \dfrac{{95040}}{{120}} = 5 \times 792 = 3960$
Hence, there are 3960 ways by which one can select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers.

Note- For such type of question just use the basic concepts of permutation and combination. Just keep in mind that ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$ where r is the number of unordered outcomes and n is the number of possibilities to choose from. Also, we should be knowing that the fundamental counting principle states that if there are p ways to do one thing, and q ways to do another thing, then there are p × q ways to do both things. It’s also known as the Multiplication or the product rule.