Question

Find the number of ways of selecting a cricket team of 11 players from 7 batsman and 6 bowlers such that there will be at least 5 bowlers in the team.

Answer 1

Hint: Here to find this, first we need to know how many possibilities can arise. Here two cases arise, such that If the selected bowlers are 5 and the selected bowlers be 6. By selecting the bowlers in the form of combination i.e. ${}^{n}{{C}_{r}}$ we can find the ways of each case by taking \[\text{number of ways = number of ways to select bowlers }\!\!\times\!\!\text{ number of ways to select batsman}\].Then we will find total ways.

Complete step-by-step answer:
In this question it is given that there are 7 batsman and 6 bowlers to make a team of 11 players in which there will be at least 5 bowlers,
Then the probabilities of selection bowlers be 5 or 6
Hence, to solve these 2 cases arises.
Case: 1 if we select 5 bowlers
Total players =11
If we take 5 bowlers then the batsman will be 6.
Now, we know that the ways in which r items can be selected among n is $^{n}{{C}_{r}}$.
Therefore, we have ways in which 5 bowlers is selected among 6 bowlers in $^{6}{{C}_{5}}$ and 6 batsman is selected among 7 batsman in ${}^{7}{{C}_{6}}$
Then, \[\text{number of ways = number of ways to select bowlers }\!\!\times\!\!\text{ number of ways to select batsman}\]
$\begin{align}
  & {{\Rightarrow }^{6}}{{C}_{5}}\times {}^{7}{{C}_{6}}=\dfrac{6!}{5!\left( 6-5 \right)!}\times \dfrac{7!}{6!\left( 7-6 \right)!} \\
 & =\dfrac{6\times 5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1\left( 1 \right)}\times \dfrac{7\times 6\times 5\times 4\times 3\times 2\times 1}{6\times 5\times 4\times 3\times 2\times 1\left( 1 \right)} \\
\end{align}$
By cancelling the common factors from numerator and denominator, we get –
$\begin{align}
  & =6\times 7 \\
 & =42\text{ ways} \\
\end{align}$
 Case: 2 if we select 6 bowlers
Total players =11
If we take 6 bowlers then the batsman be 5.
Now, we know that the ways in which r items can be selected among n is $^{n}{{C}_{r}}$.
Therefore, we have ways in which 6 bowlers is selected among 6 bowlers in $^{6}{{C}_{6}}$ and 5 batsman is selected among 7 batsman in ${}^{7}{{C}_{5}}$
Then, \[\text{number of ways = number of ways to select bowlers }\!\!\times\!\!\text{ number of ways to select batsman}\]
$\begin{align}
  & {{\Rightarrow }^{6}}{{C}_{6}}\times {}^{7}{{C}_{5}}=\dfrac{6!}{6!\left( 6-6 \right)!}\times \dfrac{7!}{5!\left( 7-5 \right)!} \\
 & =\dfrac{6\times 5\times 4\times 3\times 2\times 1}{6\times 5\times 4\times 3\times 2\times 1}\times \dfrac{7\times 6\times 5\times 4\times 3\times 2\times 1}{5\times 4\times 3\times 2\times 1\left( 2\times 1 \right)} \\
\end{align}$
By cancelling the common factors from numerator and denominator, we get –
$=1\times 7\times 3$
$=21\text{ ways}$
Therefore, the total number of ways $=42\text{ ways + 21 ways}$
$=63\text{ ways}$
Hence, the number of ways of selecting a cricket team of 11 players from 7 batsman and 6 bowlers such that there will be at least 5 bowlers in the team are 63 ways.

Note: Generally students get confused between combination & permutation. If you have to select use combination ${}^{n}{{C}_{r}}$ and if you have to arrange use permutation ${}^{n}{{P}_{r}}$ . it is very nice trick to use.
Don’t forget to consider all possibilities or else you might get the wrong answer. For example: if you missed any of the situation/case then you will get the wrong answer.