Question

An elevator starts with m passengers and stops at n floors (m $ \leqslant $ n). The probability that no two passengers leaves at the same floor is
A.$\dfrac{{^n{P_m}}}{{{m^n}}}$
B.$\dfrac{{^n{P_m}}}{{{n^m}}}$
C.$\dfrac{{^n{C_m}}}{{{m^n}}}$
D.$\dfrac{{^n{C_m}}}{{{n^m}}}$

Answer 1

Hint: In this question, m passengers leave at different floors it means we need m floors. In this question we have n floors out of which we have to choose m floors. In those m floors we have m passengers, and these passengers can be arranged in m! ways. This will be our favorable case. Our total case will be n X n X n X … m times. Let’s see how we can solve it.

Complete step-by-step answer:
We have n floors out of which we have to select m floors for ‘m’ passengers because every passenger leaves at different floors = $^n{C_m}$
Now, we have m number of passengers, so we can arrange them in m! ways.
Thus, Favorable cases =$^n{C_m}$. m! = $^n{P_m}$
Total case = n X n X n X n …... (m times) = ${n^m}$(because we need m floors)
Now, Probability = $\dfrac{{favourable\,\,cases}}{{Total\,cases}}$= $\dfrac{{^n{P_m}}}{{n.n.n....n(m\,times)}} = \dfrac{{^n{P_m}}}{{{n^m}}}$(Ans.)
Thus, the probability that no two passengers leave at the same floor is $\dfrac{{^n{P_m}}}{{{n^m}}}$.
Hence, Option B is the correct option.

Note: Probability is defined as the possibility of an event to happen is equal to the ratio of the number of favorable outcomes and the total number of outcomes. We have used this formula above to calculate probability.
Probability of event to happen P(E) = $\dfrac{{Number\,of\,favourable\,outcomes}}{{Total\,number\,of\,outcomes}}$.