Question

A student has 5 pants and 8 shirts. Find the number of ways in which he can wear the dress in different combinations.
A. ${}^{8}{{P}_{5}}$
B. ${}^{8}{{C}_{5}}$
C. $8!\times 5!$
D. $40$

Answer 1

Hint: We have to choose a dress out of 5 pants and 8 shirts. We are trying to find the number of different combinations we can take out of them. So, we try to find the number of ways he can choose 1 pant out of 5 pants and 1 shirt out of 8 shirts. The events being independent to each other we multiply the answers to find the solution of the problem.

Complete step by step answer:
The student has 5 pants and 8 shirts. To wear a complete dress, he needs 1 shirt and 1 pant. The number of pants and shirts required aren’t going to change.
So, he has to choose 1 pant out of 5 pants and 1 shirt out of 8 shirts. The pants and the shirts all are unique.
So, the number of ways a person can choose r items out of n items $\left( n\ge r \right)$ is calculated by ${}^{n}{{C}_{r}}$.
We are taking a combination of the items as in case of choosing 1 item there is no need for arrangement of the item. So, the permutation and the combination give similar results for both cases.
Here we have two cases of choosing pants and shirts. Both the events are independent to each other.
So, for every choice of the pants we can choose different ways of choosing a shirt.
So, we first choose 1 pant out of 5 pants which can be done in ${}^{5}{{C}_{1}}$ ways. Then we choose 1 shirt out of 8 shirts which can be done in ${}^{8}{{C}_{1}}$ ways.
So, the number of ways the total combination will be is ${}^{5}{{C}_{1}}\times {}^{8}{{C}_{1}}$.
Now we apply the theorem of combination as ${}^{n}{{C}_{r}}=\dfrac{n!}{\left( n-r \right)!\times r!}$. We place $r=1$ to get
${}^{n}{{C}_{1}}=\dfrac{n!}{\left( n-1 \right)!\times 1!}=n$.
Placing the values, we get ${}^{5}{{C}_{1}}\times {}^{8}{{C}_{1}}=5\times 8=40$.

So, the correct answer is “Option D”.

Note: We can also use permutation but the number of arrangements of 1 item is always 1. So, the total count stays the same for all the time. We know that ${}^{n}{{P}_{r}}={}^{n}{{C}_{r}}\times r!=\dfrac{n!\times r!}{\left( n-r \right)!\times r!}=\dfrac{n!}{\left( n-r \right)!}=n$. That’s why the solution never changes even if we try to form the permutation method.