Question

In how many ways can a cricket team be selected from a group of 25 players containing $10$ batsmen, $8$ bowlers, $5$ all-rounders, and $2$ wicketkeepers? Assume that a team of $11$ players requires $5$ batsmen, $3$ all-rounders, $2$ bowlers, and $1$ wicketkeeper.
$A) 141100$
$B) 141105$
$C) 141110$
$D) 141120$

Answer 1

Hint: In this question, First of all, try to recollect the settings of a cricket team and the number of players like batsmen, bowlers, all-rounders, and wicket keepers.
Also, we find out each group for selecting the given data.
Then multiplying each group by using the formula.
Finally, we get the required answer.

Formula used:
${}^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$

Complete step by step answer:
In the given question there are players =$25$ in which batsmen =$10$, bowlers =$8$, all-rounder =$5$ and wicket keepers $ = 2$
We have to find out the $11$ players required $5$ batsmen, $3$ all-rounder, $2$ bowlers, and $1$ wicketkeeper.
Here we have to use the combination formula that is ${}^n{C_r}$:
Now we find that the number of each group and by multiplying the term we get the require answer
So we can write it as,
Number of ways $ = $ (number of ways of choosing $5$ batsmen from $10$)$ \times $ (Number of ways of choosing $3$ all-rounders from $2$ bowlers from $8$) $ \times $ (number of ways of choosing $1$wicket keeper from $2$)
So we can write it as by using the formula
Total number of ways $ = {}^{10}{C_5} \times {}^8{C_2} \times {}^5{C_3} \times {}^2{C_1}$
Now substitute it by the formula ${}^n{C_r} = \dfrac{{n!}}{{r!(n - 1)!}}$ and we get,
$\Rightarrow \dfrac{{10!}}{{5!5!}} \times \dfrac{{8!}}{{6!2!}} \times \dfrac{{5!}}{{3!2!}} \times \dfrac{{2!}}{{1!1!}}$
Here we split the factorial term we get is
\[ = \dfrac{{10 \times 9 \times 8 \times 7 \times 6 \times 5!}}{{5! \times 5 \times 4 \times 3 \times 2 \times 1}} \times \dfrac{{8 \times 7 \times 6!}}{{6! \times 2 \times 1}} \times \dfrac{{5 \times 4 \times 3!}}{{3! \times 2 \times 1}} \times \dfrac{{2 \times 1!}}{{1! \times 1}}\]
After cancelling out the same terms from the numerator and denominator
\[ = \dfrac{{10 \times 9 \times 8 \times 7 \times 6}}{{5 \times 4 \times 3 \times 2 \times 1}} \times \dfrac{{8 \times 7}}{{2 \times 1}} \times \dfrac{{5 \times 4}}{{2 \times 1}} \times \dfrac{2}{1}\]
On multiply the numerator and denominator we get,
$ = 252 \times 28 \times 10 \times 2$
After doing multiplying the all the terms together
$ = 141120$

Hence, the correct option is $(D)$ that is $141120$.

Note:
Whenever we get this type of problem the key concept of solving is, we have to understand the laws of permutation and combination then we will be able to answer these kinds of questions.
We have used the concept of combination is
${}^n{C_r} = \dfrac{{n!}}{{r!)(n - r)!}}$$ = \dfrac{{n(n - 1)(n - 2)(n - 3)..........(n - (n - 1)!}}{{r(r - 1)(r - 2)........(r - (r - 1)!n(n - 1)(n - 2)..........(n - (n - 1)!}}$.
Students should read the question properly and also know how to apply the conditions, to ensure that the solution does not go wrong.