Question 14 Marks
$24 \ 3 \ 32 \ 1 $ A state cricket authority has to choose a team of $11$ members, to do it so the authority asks $2$ coaches of a government academy to select the team members that have experience as well as the best performers in last $15$ matches. They can make up a team of $11$ cricketers amongst $15$ possible candidates. In how many ways can the final eleven be selected from $15$ cricket players if:
$i$. Two of them being leg spinners, in how many ways can be the final eleven be selected from $15$ cricket players if one and only one leg spinner must be included? $(1)$
$ii$. If there are $6$ bowlers, $3$ wicketkeepers, and $6$ batsmen in all. In how many ways can be the final eleven be selected from $15$ cricket players if $4$ bowlers, $2$ wicketkeepers and $5$ batsmen are included. $(1)$
$iii$. In how many ways can be the final eleven be selected from $15$ cricket players if there is no restriction? $(2)$
OR
In how many ways can be the final eleven be selected from $15$ cricket players if one particular player must be included. $(2)$
$i$. Two of them being leg spinners, in how many ways can be the final eleven be selected from $15$ cricket players if one and only one leg spinner must be included? $(1)$
$ii$. If there are $6$ bowlers, $3$ wicketkeepers, and $6$ batsmen in all. In how many ways can be the final eleven be selected from $15$ cricket players if $4$ bowlers, $2$ wicketkeepers and $5$ batsmen are included. $(1)$
$iii$. In how many ways can be the final eleven be selected from $15$ cricket players if there is no restriction? $(2)$
OR
In how many ways can be the final eleven be selected from $15$ cricket players if one particular player must be included. $(2)$
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 ${ }^{13} C _{10} \times{ }^2 C _1$
${ }^n C_r=\frac{n!}{(n-r)!r!}$
$\Rightarrow{ }^{13} C_{10}=\frac{13!}{(13-10)!10!}$
$\Rightarrow{ }^{13} C_{10}=\frac{131}{31101}=\frac{13 \times 12 \times 11 \times 101}{3 \times 2 \times 1 \times 101}$
$\Rightarrow{ }^{13} C_{10}=\frac{13 \times 12 \times 11}{3 \times 2 \times 1}=13 \times 6 \times 11$
$\Rightarrow{ }^{13} C_{10}=858$
${ }^2 C _1 \times{ }^{13} C _{10}$
$\Rightarrow 2 \times 858=1716$
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 $3={ }^3 C _2={ }^3 C _1=3$
$\Rightarrow{ }^6 C_4 \times{ }^6 C_5 \times{ }^3 C_2$
$\Rightarrow 15 \times 6 \times 3=270$
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
${ }^n C_r=\frac{n!}{(n-r)|r|}$
$\Rightarrow{ }^{15} C_{11}=\frac{15!}{(15-11)!111}$
$\Rightarrow{ }^{15} C_{11}=\frac{151}{4!11!}=\frac{15 \times 14 \times 13 \times 12 \times 11!}{4 \times 3 \times 2 \times 1 \times 111}$
$\Rightarrow{ }^{15} C_{11}=\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}=15 \times 13 \times 7$
$\Rightarrow{ }^{15} C^{11}=1365$
OR
If one player must always be included, then we have to select $10$ players from $14$
${ }^n C_r=\frac{n!}{(n-r)|r|}$
$\Rightarrow{ }^{14} C_{10}=\frac{14!}{(14-10)!101}$
$\Rightarrow{ }^{14} C_{10}=\frac{14!}{4!10!}=\frac{14 \times 13 \times 12 \times 11 \times 101}{4 \times 3 \times 2 \times 1 \times 101}$
$\Rightarrow{ }^{14} C_{10}=\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1}=13 \times 11 \times 7$
$\Rightarrow{ }^{14} C_{10}=1001$
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 ${ }^{13} C _{10} \times{ }^2 C _1$
${ }^n C_r=\frac{n!}{(n-r)!r!}$
$\Rightarrow{ }^{13} C_{10}=\frac{13!}{(13-10)!10!}$
$\Rightarrow{ }^{13} C_{10}=\frac{131}{31101}=\frac{13 \times 12 \times 11 \times 101}{3 \times 2 \times 1 \times 101}$
$\Rightarrow{ }^{13} C_{10}=\frac{13 \times 12 \times 11}{3 \times 2 \times 1}=13 \times 6 \times 11$
$\Rightarrow{ }^{13} C_{10}=858$
${ }^2 C _1 \times{ }^{13} C _{10}$
$\Rightarrow 2 \times 858=1716$
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 $3={ }^3 C _2={ }^3 C _1=3$
$\Rightarrow{ }^6 C_4 \times{ }^6 C_5 \times{ }^3 C_2$
$\Rightarrow 15 \times 6 \times 3=270$
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
${ }^n C_r=\frac{n!}{(n-r)|r|}$
$\Rightarrow{ }^{15} C_{11}=\frac{15!}{(15-11)!111}$
$\Rightarrow{ }^{15} C_{11}=\frac{151}{4!11!}=\frac{15 \times 14 \times 13 \times 12 \times 11!}{4 \times 3 \times 2 \times 1 \times 111}$
$\Rightarrow{ }^{15} C_{11}=\frac{15 \times 14 \times 13 \times 12}{4 \times 3 \times 2 \times 1}=15 \times 13 \times 7$
$\Rightarrow{ }^{15} C^{11}=1365$
OR
If one player must always be included, then we have to select $10$ players from $14$
${ }^n C_r=\frac{n!}{(n-r)|r|}$
$\Rightarrow{ }^{14} C_{10}=\frac{14!}{(14-10)!101}$
$\Rightarrow{ }^{14} C_{10}=\frac{14!}{4!10!}=\frac{14 \times 13 \times 12 \times 11 \times 101}{4 \times 3 \times 2 \times 1 \times 101}$
$\Rightarrow{ }^{14} C_{10}=\frac{14 \times 13 \times 12 \times 11}{4 \times 3 \times 2 \times 1}=13 \times 11 \times 7$
$\Rightarrow{ }^{14} C_{10}=1001$
In $1001$ ways can be the final eleven be selected from $15$ cricket players if one particular player must be included.
