Question

A cricket team of eleven players is to be selected out 6 batsman, 5 bowlers, 2 all-rounders and 3 wicket keepers. Consider an all-rounder as a player who is a batsman as well as a blower.
The number selection containing exactly one wicket and at least one all-rounder is:
A. $ ^2{C_1}(2{\,^{11}}{C_9} + {\,^{11}}{C_8}) $
B. $ ^2{C_1}.{\,^2}{C_9}.{\,^{11}}{C_9} $
C. $ ^2{C_1}.{(^{12}}{C_3} + {\,^{11}}{C_3}) $
D. $ ^4{C_1}.{\,^{11}}{C_9} $

Answer 1

Hint: We will take two cases, in first we will solve to take one all-rounder and one wicket keeper. Further, we will take case 2, in case2 we will solve to take exactly one wicket keeper and at least one all-rounder. Thereafter we will add the result of both cases to get the desired result.
Formulas of combination: $ ^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $

Complete step-by-step answer:
In a cricket team total players $ = 11 $
In a cricket team total batsman $ = 6 $
In a cricket team total bowlers $ = 5 $
In a cricket team total all rounder $ = 2 $
In a cricket team total wicket keepers $ = 2 $ .
Now,
Case 1:
When one all-rounder and one wicket keeper,
So, one all-rounder out of $ 2 = {\,^2}{C_1} $
One wicket keepers out of $ 2 = {\,^2}{C_1} $
And selected $ 2 $ out of $ 11 $ players $ = {\,^{11}}{C_9} $
Then, $ ^2{C_1}{ \times ^2}{C_1} \times {\,^{11}}{C_9} $ ……(i)
Case 2:
When two all- rounder and one wicket keeper
Then two all- rounder out of $ 2 = {\,^2}{C_2} $
And one wicket keeper out of $ 2 = {\,^2}{C_1} $
Selection $ 3 $ out of $ 11 $ players $ = {\,^{11}}{C_8} $
So, $ ^2{C_2} \times {\,^2}{C_1} \times {\,^{11}}{C_8} $ ……(ii)
Now, we will add equation (i) and equation (ii),
We have
\[^2{C_1} \times {\,^2}{C_1} \times {\,^{11}}{C_9} + {\,^2}{C_2} \times {\,^2}{C_1} \times {\,^{11}}{C_8}\]
As we know that $ ^2{C_2} = 1\,\,and\,{\,^2}{C_1} = \dfrac{{2!}}{{2 - 1!1!}} $
 $ ^2{C_1} = \dfrac{{2!}}{{1!\,\,1!}} $
 $ ^2{C_1} = \dfrac{{2 \times 1}}{{1 \times 1}} $
 $ ^2{C_1} = 2 $
Then, $ 2 \times {\,^2}{C_1} \times {\,^2}{C_1} \times {\,^{11}}{C_9} + 1 \times {\,^2}{C_1} \times {\,^{11}}{C_8} $ .
Now, will take common $ ^2{C_1} $ , we will get $ = {\,^2}{C_1}(2.{\,^{11}}{C_9} + {\,^{11}}{C_8}) $
Therefore the required answer is $ = {\,^2}{C_1}(2.{\,^{11}}{C_9} + {\,^{11}}{C_8}) $
So, the correct answer is “Option A”.

Note: In case of permutation or combination problems whenever word at least or at most is given in question. Then we have to make different cases to solve the given problem, so students must be very careful regarding cases. As, if case forming is not exact in number then it will not provide the correct solution of the given problem.