Question

Among 22 cricket players, there are 3 wicket keepers and 6 bowlers. In how many ways can a team of 11 players be chosen so as to include exactly one wicketkeeper and at least 4 bowlers?

Answer 1

Hint: Since, we have to select at least $ 4 $ bowlers out of $ 6 $ given bowlers. So, we have to make different cases. In case first we will take four bowlers, in second case we take five and in third case we take six and also do the selection of other players according to the bowler chosen in different cases and hence on solving all three cases and finding their sum to have a solution to the given problem.
Formulas used: Formula of combination: $ ^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}} $

Complete step by step solution:
Total number of players = $ 22 $
Number of wicket keepers = $ 3 $
Number of bowlers = $ 6 $
Total number of players required for cricket team = $ 11 $
As, there are $ 22 $ players present out of which $ 3 $ are wicketkeeper, $ 6 $ are bowlers. Therefore, $ 13 $ are neither bowler nor wicketkeeper.
But we require only $ 11 $ out of them having $ 1 $ wicketkeeper, at least $ 4 $ bowlers and rest batsman. So, we apply the concept of combination here to get a solution to this problem.
Since, as we require at least $ 4 $ bowlers. Therefore, different cases will be formed. Which are as follows:
Case (i)
In this case we choose $ 4\,\,bowlers,\,1\,wicketkeeper\,\,and\,\,6\,\,batsman $ out of $ 6\,\,bowlers,\,\,3\,\,wicketkeepers\,\,and\,\,13\,\,batsman $ . This can be done as:

 $\Rightarrow {^6}{C_4}{ \times ^3}{C_1}{ \times ^{13}}{C_6} $
\[
   = \dfrac{{6!}}{{4!2!}} \times \dfrac{{3!}}{{2!1!}} \times \dfrac{{13!}}{{6!7!}} \\
   = \dfrac{{6 \times 5 \times 4!}}{{4! \times 2 \times 1}} \times \dfrac{{3 \times 2!}}{{2!}} \times \dfrac{{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7!}}{{6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 7!}} \\
   = 15 \times 3 \times 1716 \\
   = 77220 \\
 \]

Case (ii) .
In this case we choose $ 5\,\,bowlers,\,1\,wicketkeeper\,\,and\,\,5\,\,batsman $ out of $ 6\,\,bowlers,\,\,3\,\,wicketkeepers\,\,and\,\,13\,\,batsman $ . This can be done as:
 $\Rightarrow {^6}{C_5}{ \times ^3}{C_1}{ \times ^{13}}{C_5} $
 $
   = \dfrac{{6!}}{{5!1!}} \times \dfrac{{3!}}{{1! \times 2!}} \times \dfrac{{13!}}{{5!8!}} \\
   = \dfrac{{6 \times 5!}}{{5!}} \times \dfrac{{3 \times 2!}}{{2!}} \times \dfrac{{13 \times 12 \times 11 \times 10 \times 9 \times 8!}}{{8! \times 5 \times 4 \times 3 \times 2 \times 1}} \\
   = 6 \times 3 \times 1287 \\
   = 23166 \\
  $
Case (iii)
In this case we choose $ 6\,\,bowlers,\,1\,wicketkeeper\,\,and\,\,4\,\,batsman $ out of $ 6\,\,bowlers,\,\,3\,\,wicketkeepers\,\,and\,\,13\,\,batsman $ . This can be done as:
 $\Rightarrow {^6}{C_6}{ \times ^3}{C_1}{ \times ^{13}}{C_4} $
 $
   = \dfrac{{6!}}{{6!0!}} \times \dfrac{{3!}}{{2!1!}} \times \dfrac{{13!}}{{4!9!}} \\
   = 1 \times \dfrac{{3 \times 2!}}{{2!}} \times \dfrac{{13 \times 12 \times 11 \times 10 \times 9!}}{{4 \times 3 \times 2 \times 1 \times 9!}} \\
   = 1 \times 3 \times 715 \\
   = 2145 \;
  $
Hence, from above we that total ways of selecting a wicketkeeper and at least four bowlers from given twenty two players are = $ 77220 + 23166 + 2145 $
 $ = 102531\,\,ways $

Note: If there are more players available but less required. So, we apply the concept of combination instead for permutation. Also, if there is a word at least and at most in problem. Then we have to make different cases as per given condition and then finding their sum after solving different cases formed as per or at least or at most gives out the required solution of the given problem.