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Question: How many cricket teams with members of eleven each can be formed from 15 persons if the captain is i...

How many cricket teams with members of eleven each can be formed from 15 persons if the captain is included in every team?
A. 364
B. 1365
C. 1001
D. 1000

Explanation

Solution

Hint: In this question, we have to find the number of ways of choosing 11 members out of 15 members and out of those 11 members, the captain is always included. So, we have to choose 10 out of 14 members. We should know that when we have to choose rr elements out of nn elements, then we have to apply the formula of combination, that is, nCr=n!r!(nr)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}.

Complete step-by-step answer:

In this question, we have to find the number of ways to form a team of 11 members out of 15 members, in which the captain is always included. So, basically, we have to choose 10 out of 14 members. We know that when we have to choose rr items out of nn items, then we have to apply the formula of combination, that is, nCr=n!r!(nr)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}. So, to choose 10 out of 14 members, we will consider n=14n=14 and r=10r=10. So, by applying it in the formula, we get,
14C10=14!10!(1410)!^{14}{{C}_{10}}=\dfrac{14!}{10!\left( 14-10 \right)!}
Which can be further simplified and written as,
14C10=14!10!4!^{14}{{C}_{10}}=\dfrac{14!}{10!4!}
We know that for any positive integer aa, a!=a×(a1)×(a2)×(a3)1a!=a\times \left( a-1 \right)\times \left( a-2 \right)\times \left( a-3 \right)\ldots 1. So, we can write it as,
14C10=14×13×12×11××110×9×8×7××1×4×3×2×1^{14}{{C}_{10}}=\dfrac{14\times 13\times 12\times 11\times \ldots \ldots \times 1}{10\times 9\times 8\times 7\times \ldots \ldots \times 1\times 4\times 3\times 2\times 1}
Now, we can see that 10×9×8×7××110\times 9\times 8\times 7\times \ldots \ldots \times 1 is common in both the numerator and the denominator. So, by cancelling the like terms, we get,
14C10=14×13×12×114×3×2×1 14C10=7×13×11 14C10=1001 \begin{aligned} & ^{14}{{C}_{10}}=\dfrac{14\times 13\times 12\times 11}{4\times 3\times 2\times 1} \\\ & {{\Rightarrow }^{14}}{{C}_{10}}=7\times 13\times 11 \\\ & {{\Rightarrow }^{14}}{{C}_{10}}=1001 \\\ \end{aligned}
So, the number of ways of choosing a team of 11 members out of 15 members such that the captain is always included is 1001.
Hence the correct answer is option C.

Note: The possible mistake that the students can make in this question is by considering n=15n=15 and r=11r=11 in the formula, nCr=n!r!(nr)!^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}, which is wrong as the captain of the team is always included in the team, so we have to choose only 10 out of 14 members.