Question

A cricket team of 11 players is to be formed from 20 players including 6 bowlers and 3 wicket keepers. The number of ways in which a team can be formed having exactly 4 bowlers and 2 wicket keepers is:
A. $20790$
B. $6930$
C. $10790$
D. $360$

Answer 1

We have to find the number of ways in which we can select a team of 11 players from the 20 players keeping in mind the number of bowlers and keepers which is given. So, we can select the specific number of bowlers from the total number of bowlers . The number of ways in which a team can be formed having exactly 4 bowlers is ${}^{6}{{C}_{4}}$ .Similarly, the number of ways in which a team can be formed having exactly 2 wicket keepers is ${}^{3}{{C}_{2}}$ . We can select the remaining number of players that are to be selected to complete the team of 11 players in ${}^{20-9}{{C}_{11-6}}={}^{11}{{C}_{5}}$ ways. Hence, the number of ways in which a team of 11 can be formed from 20 having exactly 4 bowlers and 2 wicket keepers is ${}^{6}{{C}_{4}}\cdot {}^{3}{{C}_{2}}\cdot {}^{11}{{C}_{5}}$ . Use $^{n}C{}_{r}=\dfrac{n!}{r!\left( n-r \right)!}$ to solve this.

Complete step by step answer:
We have to find the number of ways in which we can select a team of 11 players having exactly 4 bowlers and 2 wicket keepers from the 20 players including 6 bowlers and 3 wicket keepers .
We need to have exactly 4 bowlers given that there are 6 bowlers. The number of ways in which these are arranged is shown below
${}^{6}{{C}_{4}}$ .
We also need exactly 2 wicket keepers 3 wicket keepers . The number of ways in which these are arranged is shown below
${}^{3}{{C}_{2}}$ .
Now, the number of ways in which the remaining of the 20 players joins the team. This can be shown as
${}^{20-9}{{C}_{11-6}}={}^{11}{{C}_{5}}$
Hence, the number of ways in which a team of 11 can be formed having exactly 4 bowlers and 2 wicket keepers is
${}^{6}{{C}_{4}}\cdot {}^{3}{{C}_{2}}\cdot {}^{11}{{C}_{5}}...(i)$
We know that $^{n}C{}_{r}=\dfrac{n!}{r!\left( n-r \right)!}$
Hence, ${}^{6}{{C}_{4}}$ can be founds as follows:
${}^{6}{{C}_{4}}=\dfrac{6!}{4!(6-4)!}$
Let us simplify this.
${}^{6}{{C}_{4}}=\dfrac{6!}{4!2!}$
Now, $6!$ can be written as
${}^{6}{{C}_{4}}=\dfrac{6\times 5\times 4!}{4!2!}$
Cancelling, $4!$ from numerator and denominator, we will get
${}^{6}{{C}_{4}}=\dfrac{6\times 5}{2!}$
Solving, this gives
${}^{6}{{C}_{4}}=\dfrac{30}{2}=15...(a)$
Let us consider ${}^{3}{{C}_{2}}$ . This can be founds as follows:
${}^{3}{{C}_{2}}=\dfrac{3!}{2!(3-2)!}$
Let us simplify this.
${}^{3}{{C}_{2}}=\dfrac{3!}{2!1!}$
Now, $3!$ can be written as
${}^{3}{{C}_{2}}=\dfrac{3\times 2!}{2!}$
Cancelling, $2!$ from numerator and denominator, we will get
${}^{3}{{C}_{2}}=3...(b)$
Let us consider ${}^{11}{{C}_{5}}$ . This can be founds as follows:
${}^{11}{{C}_{5}}=\dfrac{11!}{5!(11-5)!}$
Let us simplify this.
${}^{11}{{C}_{5}}=\dfrac{11!}{5!6!}$
Now, $11!$ can be written as
${}^{11}{{C}_{5}}=\dfrac{11\times 10\times 9\times 8\times 7\times 6!}{5!6!}$
Cancelling, $6!$ from numerator and denominator, we will get
${}^{11}{{C}_{5}}=\dfrac{11\times 10\times 9\times 8\times 7}{5!}$
Now let us expand $5!$ .
${}^{11}{{C}_{5}}=\dfrac{11\times 10\times 9\times 8\times 7}{5\times 4\times 3\times 2\times 1}$
Cancelling the terms, we will get
${}^{11}{{C}_{5}}=11\times 2\times 3\times 7=462...(c)$
Now, let us substitute $(a),(b),(c)$ in $(i)$ . We will get
${}^{6}{{C}_{4}}\cdot {}^{3}{{C}_{2}}\cdot {}^{11}{{C}_{5}}=15\times 3\times 462=20790$
Hence, the total number of ways is 20790.

So, the correct answer is “Option A”.

Note: The students can make an error if they don’t know about the fundamental theorem because knowing this formula is very important to solve this question and without this theorem one could not get to the correct answer. Now, for writing the expression, ${}^{6}{{C}_{4}}\cdot {}^{3}{{C}_{2}}\cdot {}^{11}{{C}_{5}}$ , we would require the fundamental theorem which states that for finding chances of any number of events together, we just multiply their individual chances and thus we get the required answer.
Do not get confused with the equation of combination and permutation. Equation of permutation is ${}^{n}{{P}_{r}}=\dfrac{n!}{(n-r)!}$ .