Question

The Indian Cricket team consists of \[16\] players. It includes \[2\] wicket keepers and \[5\] bowlers. In how many ways can a cricket eleven be selected if we have to select \[1\] wicket keeper and at least \[4\] bowlers?
A. \[1092\]
B. \[1024\]
C. \[2550\]
D. \[3425\]

Answer 1

Hint: We are asked the no of ways in how we can form a cricket team of eleven players, with one wicket keeper and at least four bowlers. Check how many cases can be formed in selecting at least four bowlers, then calculate the no of ways for each case separately using combination formula and then combine them to find the required no of ways.

Complete step-by-step answer:
Given, total no of players \[T = 16\]
No of wicket keepers, \[W = 2\]
No of bowlers, \[B = 5\]
Remaining players, \[R = 16 - (2 + 5) = 16 - 7 = 9\]
We are asked in how many ways a team of \[11\] can be formed with \[1\] wicket keeper and at least \[4\] bowlers. For this we will use the combination formula which gives the no of ways an item can be selected,
 \[^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}\] ,where \[n\] is the total no of items and \[r\] is the required no of items.
There can be two cases,
Case 1: When there are \[1\] wicket keeper, \[4\] bowlers then remaining players needed will be \[(11 - (1 + 4)) = 6\]
Using combination formula,
No of ways \[1\] wicket keeper can be selected from \[2\] wicket keepers is
 \[^2{C_1} = \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}} = 2\]
No of ways \[4\] bowlers can be selected from \[5\] bowlers is
 \[^5{C_4} = \dfrac{{5!}}{{4!\left( {5 - 4} \right)!}} = 5\]
No of ways remaining \[6\] can be selected from \[9\] is
 \[^9{C_6} = \dfrac{{9!}}{{6!\left( {9 - 6} \right)!}} = \dfrac{{9 \times 8 \times 7}}{{3 \times 2}} = 84\]
Therefore, total no of ways for case 1 will be,
 \[{\text{case}}\,1{ = ^2}{C_1}^5{C_4}^9{C_6} = 2 \times 5 \times 84 = 840\]

Case 2: When there are \[1\] wicket keeper, \[5\] bowlers then remaining players needed will be \[(11 - (1 + 5)) = 5\]
Using combination formula,
No of ways \[1\] wicket keeper can be selected from \[2\] wicket keepers is
 \[^2{C_1} = \dfrac{{2!}}{{1!\left( {2 - 1} \right)!}} = 2\]
No of ways \[5\] bowlers can be selected from \[5\] bowlers is
 \[^5{C_5} = \dfrac{{5!}}{{5!\left( {5 - 5} \right)!}} = 1\]
No of ways remaining \[5\] can be selected from \[9\] is
 \[^9{C_5} = \dfrac{{9!}}{{5!\left( {9 - 5} \right)!}} = \dfrac{{9 \times 8 \times 7 \times 6}}{{4 \times 3 \times 2}} = 126\]
Therefore, total no of ways for case 2 will be,
 \[{\text{case}}\,2{ = ^2}{C_1}^5{C_5}^9{C_5} = 2 \times 1 \times 126 = 252\]
Total no of ways will be, addition of case 1 and case 2,
 \[{\text{Total}} = {\text{case}}\,{\text{1}} + {\text{case}}\,{\text{2}} \\
   \Rightarrow {\text{Total}} = 840 + 252 \\
   \Rightarrow {\text{Total}} = 1092 \]
So, the correct answer is “Option A”.

Note: There are two important terms in case of selecting or arranging a set of numbers, these are permutation and combination. Permutation refers to arrangement or rearrangement of a set of numbers while combination is selecting the no of ways an item can be selected from a set of items. Students usually get confused between these two terms, so always remember their difference