Question

In a building, an elevator starts from the ground floor (that is, 0th floor) with 10 passengers and stops at every floor until the 15th floor which is the topmost floor. No new passengers enter the elevator on any floor from the 1st through the 15th floor. If all the passengers got off the elevator then the probability that at most one passenger got off at each floor is

• A. $\frac{15!}{5! \times 15^{10}}$
• B. $\frac{10!}{15^{10}}$
• C. $\frac{15!}{10^{15}}$
• D. $\frac{15!}{5! \times 10^{15}}$

Hint:

This problem is identical to distributing $n$ distinct items into $N$ distinct bins with a maximum capacity of 1.
Using the permutation notation ${}^N P_n / N^n$ makes finding the answer straightforward.

Answer 1

The Correct Option is A

Step 1: Frame the experiment.
Ten distinct passengers each step out at one of the floors 1 through 15. We want the chance that no two of them leave at the same floor, that is, at most one per floor.

Step 2: Count all the possible ways to get off.
Each passenger independently has $15$ floor choices, so the total number of outcomes is $15^{10}$. This is the denominator.

Step 3: Count the favourable ways.
We need all 10 to pick different floors out of 15. The first passenger has $15$ choices, the next $14$, and so on down to $15 - 9 = 6$. That product is $15 \times 14 \times \dots \times 6$.

Step 4: Write that product as factorials.
$15 \times 14 \times \dots \times 6 = \frac{15!}{5!}$, since the leftover bottom part $5 \times 4 \times \dots \times 1 = 5!$ is what we divide out.

Step 5: Form the probability.
Probability $= \frac{\text{favourable}}{\text{total}} = \frac{15!/5!}{15^{10}} = \frac{15!}{5! \times 15^{10}}$.

Step 6: Confirm with the options.
This matches the first choice exactly.
\[ \boxed{\dfrac{15!}{5! \times 15^{10}}} \]