Answer
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Hint: We solve this problem by dividing the question in two parts. First part consists of 5 bowlers and 6 batsmen and the second part consists of 6 bowlers and 5 batsmen because at least 5 bowlers mean 5 or more bowlers. Then we use the permutations and combinations to find the required result. The number of ways of selecting \['r'\] members from \['n'\] members is given as \[{}^{n}{{C}_{r}}\] where,
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Complete step-by-step answer:
We are given that there are a total of 15 players of which 6 are bowlers.
So, we can say that there are 9 batsmen in the 15 players.
We are asked to find the number of ways of selecting 11 players consisting of at least 5 bowlers.
Let us divide this into two parts.
(i) Taking 11 players having 5 bowlers.
We know that the number of ways of selecting \['r'\] members from \['n'\] members is given as \[{}^{n}{{C}_{r}}\] where,
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
By using the above formula, we get the number of ways of selecting 5 bowlers from 6 bowlers as
\[\begin{align}
& \Rightarrow {}^{6}{{C}_{5}}=\dfrac{6!}{5!\left( 6-5 \right)!} \\
& \Rightarrow {}^{6}{{C}_{5}}=6 \\
\end{align}\]
Here, we can see that we have already selected 5 players for a team of 11 players.
Now, let us find the number of ways of selecting 6 batsmen from 9 batsmen
By using the above formula we get the number of ways of selecting 6 batsmen from 9 batsmen as
\[\begin{align}
& \Rightarrow {}^{9}{{C}_{6}}=\dfrac{9!}{6!\left( 9-6 \right)!} \\
& \Rightarrow {}^{9}{{C}_{6}}=\dfrac{9\times 8\times 7}{3\times 2}=84 \\
\end{align}\]
Let us assume that the total number of ways of selecting 5 bowlers and 6 batsmen as \['x'\]
Here we can see that while combining both the numbers that is the number of ways of selecting 5 bowlers and 6 batsmen is a type of permutation.
By using the permutations we get the total number of ways of selecting 5 bowlers and 6 batsmen as
\[\begin{align}
& \Rightarrow x=6\times 84 \\
& \Rightarrow x=504 \\
\end{align}\]
(ii) Taking 11 players having 6 bowlers.
We know that the number of ways of selecting \['r'\] members from \['n'\] members is given as \[{}^{n}{{C}_{r}}\] where,
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
By using the above formula, we get the number of ways of selecting 6 bowlers from 6 bowlers as
\[\begin{align}
& \Rightarrow {}^{6}{{C}_{6}}=\dfrac{6!}{6!\left( 6-6 \right)!} \\
& \Rightarrow {}^{6}{{C}_{6}}=1 \\
\end{align}\]
Here, we can see that we have already selected 6 players for a team of 11 players.
Now, let us find the number of ways of selecting 5 batsmen from 9 batsmen
By using the above formula we get the number of ways of selecting 5 batsmen from 9 batsmen as
\[\begin{align}
& \Rightarrow {}^{9}{{C}_{5}}=\dfrac{9!}{5!\left( 9-5 \right)!} \\
& \Rightarrow {}^{9}{{C}_{5}}=\dfrac{9\times 8\times 7\times 6}{4\times 3\times 2}=126 \\
\end{align}\]
Let us assume that the total number of ways of selecting 6 bowlers and 5 batsmen as \['y'\]
Here we can see that while combining both the numbers that is the number of ways of selecting 6 bowlers and 5 batsmen is a type of permutation.
By using the permutations we get the total number of ways of selecting 6 bowlers and 5 batsmen as
\[\begin{align}
& \Rightarrow x=1\times 126 \\
& \Rightarrow x=126 \\
\end{align}\]
Let us assume that the total number of ways of selecting at least 5 bowlers as \['N'\] which is the combination of selecting 5 bowlers and 6 bowlers
By using the above condition we get
\[\begin{align}
& \Rightarrow N=x+y \\
& \Rightarrow N=504+126 \\
& \Rightarrow N=630 \\
\end{align}\]
Therefore, there are 630 ways of selecting at least 5 bowlers to form a team of 11 players from 15 players.
So, the correct answer is “Option (a)”.
Note: Students may make mistakes at permutations and combinations.
The number of ways of selecting 5 bowlers and 6 batsmen to form a team of 11 players is a permutation type. So, we get
\[\Rightarrow x=6\times 84\]
Whereas, selecting at least 5 bowlers is a combination of selecting 5 bowlers and selecting 6 bowlers. So, we get
\[\Rightarrow N=x+y\]
Students may make mistakes in understanding the permutations concept and combinations concept.
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
Complete step-by-step answer:
We are given that there are a total of 15 players of which 6 are bowlers.
So, we can say that there are 9 batsmen in the 15 players.
We are asked to find the number of ways of selecting 11 players consisting of at least 5 bowlers.
Let us divide this into two parts.
(i) Taking 11 players having 5 bowlers.
We know that the number of ways of selecting \['r'\] members from \['n'\] members is given as \[{}^{n}{{C}_{r}}\] where,
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
By using the above formula, we get the number of ways of selecting 5 bowlers from 6 bowlers as
\[\begin{align}
& \Rightarrow {}^{6}{{C}_{5}}=\dfrac{6!}{5!\left( 6-5 \right)!} \\
& \Rightarrow {}^{6}{{C}_{5}}=6 \\
\end{align}\]
Here, we can see that we have already selected 5 players for a team of 11 players.
Now, let us find the number of ways of selecting 6 batsmen from 9 batsmen
By using the above formula we get the number of ways of selecting 6 batsmen from 9 batsmen as
\[\begin{align}
& \Rightarrow {}^{9}{{C}_{6}}=\dfrac{9!}{6!\left( 9-6 \right)!} \\
& \Rightarrow {}^{9}{{C}_{6}}=\dfrac{9\times 8\times 7}{3\times 2}=84 \\
\end{align}\]
Let us assume that the total number of ways of selecting 5 bowlers and 6 batsmen as \['x'\]
Here we can see that while combining both the numbers that is the number of ways of selecting 5 bowlers and 6 batsmen is a type of permutation.
By using the permutations we get the total number of ways of selecting 5 bowlers and 6 batsmen as
\[\begin{align}
& \Rightarrow x=6\times 84 \\
& \Rightarrow x=504 \\
\end{align}\]
(ii) Taking 11 players having 6 bowlers.
We know that the number of ways of selecting \['r'\] members from \['n'\] members is given as \[{}^{n}{{C}_{r}}\] where,
\[{}^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}\]
By using the above formula, we get the number of ways of selecting 6 bowlers from 6 bowlers as
\[\begin{align}
& \Rightarrow {}^{6}{{C}_{6}}=\dfrac{6!}{6!\left( 6-6 \right)!} \\
& \Rightarrow {}^{6}{{C}_{6}}=1 \\
\end{align}\]
Here, we can see that we have already selected 6 players for a team of 11 players.
Now, let us find the number of ways of selecting 5 batsmen from 9 batsmen
By using the above formula we get the number of ways of selecting 5 batsmen from 9 batsmen as
\[\begin{align}
& \Rightarrow {}^{9}{{C}_{5}}=\dfrac{9!}{5!\left( 9-5 \right)!} \\
& \Rightarrow {}^{9}{{C}_{5}}=\dfrac{9\times 8\times 7\times 6}{4\times 3\times 2}=126 \\
\end{align}\]
Let us assume that the total number of ways of selecting 6 bowlers and 5 batsmen as \['y'\]
Here we can see that while combining both the numbers that is the number of ways of selecting 6 bowlers and 5 batsmen is a type of permutation.
By using the permutations we get the total number of ways of selecting 6 bowlers and 5 batsmen as
\[\begin{align}
& \Rightarrow x=1\times 126 \\
& \Rightarrow x=126 \\
\end{align}\]
Let us assume that the total number of ways of selecting at least 5 bowlers as \['N'\] which is the combination of selecting 5 bowlers and 6 bowlers
By using the above condition we get
\[\begin{align}
& \Rightarrow N=x+y \\
& \Rightarrow N=504+126 \\
& \Rightarrow N=630 \\
\end{align}\]
Therefore, there are 630 ways of selecting at least 5 bowlers to form a team of 11 players from 15 players.
So, the correct answer is “Option (a)”.
Note: Students may make mistakes at permutations and combinations.
The number of ways of selecting 5 bowlers and 6 batsmen to form a team of 11 players is a permutation type. So, we get
\[\Rightarrow x=6\times 84\]
Whereas, selecting at least 5 bowlers is a combination of selecting 5 bowlers and selecting 6 bowlers. So, we get
\[\Rightarrow N=x+y\]
Students may make mistakes in understanding the permutations concept and combinations concept.
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