Question

A boy has $8$ trousers and $10$ shirts. In how many ways can he select a shirt and a trouser?
A. $80$
B. $8!$ $ \times $ $10!$
C. $64$
D. $8{!^2}$

Answer 1

Hint:In question they given 8 trousers and 10 shirts then,the product of choosing 1 trouser from 8 trousers and 1 shirt from 10 shirts gives the number of ways of selecting shirt and trouser which is the required answer.

Complete step-by-step answer:
Here the question is saying that a boy has $8$ trousers and $10$ shirts and if he wants one trouser and one shirt, in how many ways can he select it.
So as we know that if we have $n$ number of things and we have to choose $m$ number of things from it, it can be done in \[{}^n{C_m}\] ways which means$\dfrac{{n!}}{{m!\left( {m - n} \right)!}}$.
Here the boy is choosing $1$ trouser out of $10$ trousers. So he can do it in ${}^8{C_1}$ ways.
Which is$\dfrac{{8!}}{{1!7!}}$ $ = 8$.
Boy is also choosing $1$ shirts out of $10$ shirts.
So number of ways of doing this$ = {}^{10}{C_1} = \dfrac{{10!}}{{1!\left( {10 - 1} \right)!}} = \dfrac{{10!}}{{1!9!}} = 10$.
Hence the number of ways he can select a shirt and a trouser out of $8$ trousers and $10$ shirts is
$ = $ Number of ways to choose $1$ shirt $ \times $ number of ways to choose $1$ trouser.
$ = $ $8 \times 10 = 80$
So there are $80$ ways to select one shirt and one trouser.

So, the correct answer is “Option A”.

Note:We can choose $m$ number of things from $n$ number of things in \[{}^n{C_m}\] ways which means $\dfrac{{n!}}{{m!\left( {m - n} \right)!}}$. For example: we can choose $2$ objects from $10$ objects in ${}^{10}{C_2}$ ways which is $ = \dfrac{{10!}}{{2!8!}}$$ = 45$ ways.