Question 14 Marks
Answer
View full question & answer→i. Two of them being leg spinners, one and only one leg spinner must be included
Let's first find out possible ways to select players which are not leg spinner
There are two leg spinners out of 15 and one players must be leg spinner.
So, we have to select 10 players out of 13
Total possible ways to select 11 players out of 15 out of which one must be leg spinner out of 2 are ${ }^{13} C _{10} \times{ }^2 C _1$
${ }^n C_r=\frac{n!}{(n-r)!r!}$
$\Rightarrow{ }^{13} C_{10}=\frac{13!}{(13-10)!10!}$
$\begin{array}{l}\Rightarrow{ }^{13} C_{10}=\frac{131}{31101}=\frac{13 \times 12 \times 11 \times 101}{3 \times 2 \times 1 \times 101} \\ \Rightarrow{ }^{13} C_{10}=\frac{13 \times 12 \times 11}{3 \times 2 \times 1}=13 \times 6 \times 11 \\ \Rightarrow{ }^{13} C_{10}=858\end{array}$
${ }^2 C _1 \times{ }^{13} C _{10}$
$\Rightarrow 2 \times 858=1716$
Total possible ways to select 11 players out of 15 out of which one must be leg spinner out of 2 = 1716
ii. number of ways of selecting 4 bowlers out of $6={ }^6 C _4$
$\Rightarrow{ }^6 C_4=\frac{61}{(6-4)|4|}=\frac{61}{214!}=\frac{6 \times 5 \times 4!}{2 \times 1 \times 4!}=15$
number of ways of selecting 5 batsmen out of $6={ }^6 C _5=6$
number of ways of selecting 2 wicket keepers out of $3={ }^3 C _2={ }^3 C _1=3$
$\begin{array}{l}\Rightarrow{ }^6 C_4 \times{ }^6 C_5 \times{ }^3 C_2 \\
\Rightarrow 15 \times 6 \times 3=270\end{array}$
Total ways to select 4 bowlers, 2 wicketkeepers and 5 batsmen out of 6 bowlers, 3 wicketkeepers, and 6 batsmen in all are 270.
iii. Here, we have to select 11 players out of 15 and there are no restrictions and here the order of the players doesn't matter. So, we will here apply combination
${ }^n C_r=\frac{n!}{(n-r)|r|}$
$\begin{array}{l}\Rightarrow{ }^{15} C_{11}=\frac{15!}{(15-11)!111} \\ \Rightarrow{ }^{15} C_{11}=\frac{151}{4!11!}=\frac{15 \times 14 \times 13 \times 12 \times 11!}{4 \times 3 \times 2 \times 1 \times 111} \\ \Rightarrow{ }^{15} C_{11}=\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}=15 \times 13 \times 7 \\ \Rightarrow{ }^{15} C^{11}=1365\end{array}$
OR
If one player must always be included, then we have to select 10 players from 14
${ }^n C_r=\frac{n!}{(n-r)|r|}$
$\begin{array}{l}\Rightarrow{ }^{14} C_{10}=\frac{14!}{(14-10)!101} \\ \Rightarrow{ }^{14} C_{10}=\frac{14!}{4!10!}=\frac{14 \times 13 \times 12 \times 11 \times 101}{4 \times 3 \times 2 \times 1 \times 101} \\ \Rightarrow{ }^{14} C_{10}=\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1}=13 \times 11 \times 7 \\ \Rightarrow{ }^{14} C_{10}=1001\end{array}$
In 1001 ways can be the final eleven be selected from 15 cricket players if one particular player must be included.
Let's first find out possible ways to select players which are not leg spinner
There are two leg spinners out of 15 and one players must be leg spinner.
So, we have to select 10 players out of 13
Total possible ways to select 11 players out of 15 out of which one must be leg spinner out of 2 are ${ }^{13} C _{10} \times{ }^2 C _1$
${ }^n C_r=\frac{n!}{(n-r)!r!}$
$\Rightarrow{ }^{13} C_{10}=\frac{13!}{(13-10)!10!}$
$\begin{array}{l}\Rightarrow{ }^{13} C_{10}=\frac{131}{31101}=\frac{13 \times 12 \times 11 \times 101}{3 \times 2 \times 1 \times 101} \\ \Rightarrow{ }^{13} C_{10}=\frac{13 \times 12 \times 11}{3 \times 2 \times 1}=13 \times 6 \times 11 \\ \Rightarrow{ }^{13} C_{10}=858\end{array}$
${ }^2 C _1 \times{ }^{13} C _{10}$
$\Rightarrow 2 \times 858=1716$
Total possible ways to select 11 players out of 15 out of which one must be leg spinner out of 2 = 1716
ii. number of ways of selecting 4 bowlers out of $6={ }^6 C _4$
$\Rightarrow{ }^6 C_4=\frac{61}{(6-4)|4|}=\frac{61}{214!}=\frac{6 \times 5 \times 4!}{2 \times 1 \times 4!}=15$
number of ways of selecting 5 batsmen out of $6={ }^6 C _5=6$
number of ways of selecting 2 wicket keepers out of $3={ }^3 C _2={ }^3 C _1=3$
$\begin{array}{l}\Rightarrow{ }^6 C_4 \times{ }^6 C_5 \times{ }^3 C_2 \\
\Rightarrow 15 \times 6 \times 3=270\end{array}$
Total ways to select 4 bowlers, 2 wicketkeepers and 5 batsmen out of 6 bowlers, 3 wicketkeepers, and 6 batsmen in all are 270.
iii. Here, we have to select 11 players out of 15 and there are no restrictions and here the order of the players doesn't matter. So, we will here apply combination
${ }^n C_r=\frac{n!}{(n-r)|r|}$
$\begin{array}{l}\Rightarrow{ }^{15} C_{11}=\frac{15!}{(15-11)!111} \\ \Rightarrow{ }^{15} C_{11}=\frac{151}{4!11!}=\frac{15 \times 14 \times 13 \times 12 \times 11!}{4 \times 3 \times 2 \times 1 \times 111} \\ \Rightarrow{ }^{15} C_{11}=\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}=15 \times 13 \times 7 \\ \Rightarrow{ }^{15} C^{11}=1365\end{array}$
OR
If one player must always be included, then we have to select 10 players from 14
${ }^n C_r=\frac{n!}{(n-r)|r|}$
$\begin{array}{l}\Rightarrow{ }^{14} C_{10}=\frac{14!}{(14-10)!101} \\ \Rightarrow{ }^{14} C_{10}=\frac{14!}{4!10!}=\frac{14 \times 13 \times 12 \times 11 \times 101}{4 \times 3 \times 2 \times 1 \times 101} \\ \Rightarrow{ }^{14} C_{10}=\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1}=13 \times 11 \times 7 \\ \Rightarrow{ }^{14} C_{10}=1001\end{array}$
In 1001 ways can be the final eleven be selected from 15 cricket players if one particular player must be included.
