Question

There are 5 doors to a lecture hall. The number of ways that a student can enter the hall and leave it by a different door is
A. 20
B. 16
C. 19
D. 21

Answer 1

Hint: We have a lecture hall with 5 doors. The number of ways of selecting r objects from n objects is given by ${}^n{{\text{C}}_r}{\text{ = }}\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. So the number of ways a student can enter through anyone out of the 5 doors is given by ${}^5{{\text{C}}_1}$. And the number of ways a student can leave the classroom through any of the 4 doors left is given by ${}^4{{\text{C}}_1}$. The total number of ways can be calculated by taking the product of the number of ways of entering and the number of ways of leaving the classroom.

Complete step by step answer:

We have 5 doors. A student can enter through any of the 5 doors.
We know that number of ways of selecting r objects from n objects is given by ${}^n{{\text{C}}_r}$, where ${}^n{{\text{C}}_r}{\text{ = }}\dfrac{{n!}}{{r!\left( {n - r} \right)!}}$
Using the above relation, we can write the number of ways a student can enter to the lecture hall by ${}^{\text{5}}{{\text{C}}_{\text{1}}}{\text{ = }}\dfrac{{5!}}{{1!\left( {5 - 1} \right)!}}{\text{ = }}\dfrac{{5 \times 4!}}{{4!}}{\text{ = 5}}$.
As a student who entered in the hall can’t leave through the door he entered. So, he can use any one door out of the four doors left. So we get the number of ways that the student can leave the hall is${}^{\text{4}}{{\text{C}}_{\text{1}}}{\text{ = }}\dfrac{{4!}}{{1!\left( {4 - 1} \right)!}} = \dfrac{{4 \times 3!}}{{3!}}{\text{ = 4}}$.
The number of ways that a student can enter the hall and leave it by a different door is given by the product ${}^{\text{5}}{{\text{C}}_{\text{1}}}{\text{ $\times$ }}{}^{\text{4}}{{\text{C}}_{\text{1}}}{\text{ = 5 $\times$ 4 = 20}}$.
There 20 ways that a student can enter a hall with 5 doors and leave it by a different door.
Therefore, the correct answer is option A.

Note: The concept of permutations and combinations are used. To find the total number of ways we must take the product not the sum. We can also solve the problem without using the concept of combinations as follows. There are 5 doors to enter and only 4 to leave, so we can directly multiply to get the total number of ways as 20.