Hello, fatcarl!

Five people enter an elevator at the ground floor af an 8 story builiding.

Assuming that each person is equally likely to get off at any one of the 7 floors above ground,

what is the probablity that exactly 2 people will get off at the same floor?

(ie 2 people exit on one of the floors and the other 3 people exit alone, each on a different floor)

I'll assume that the five people are distinguishable (i.e., have different names).

Person "A" has 7 choices of floors, person "B" has 7 choices of floors, person "C" has 7 choices, etc.

. . Hence, there are \(\displaystyle 7^5\,=\,16,807\) possible ways they could exit.

First, select the two that will exit together.

.There are \(\displaystyle C(5,2)\,=\,\frac{5!}{2!3!}\,=\,10\) possible pairs.

. . Duct-tape the pair together.

.Then we have 4 "people" to consider.

. . These 'four' can exit in:

.\(\displaystyle P(7,4)\,=\,\frac{7!}{3!}\,=\,840\) ways.

Hence, they can exit in: \(\displaystyle 10\,\times\,840\:=\:8400\) ways.

The probability is:

.\(\displaystyle \frac{8400}{16807}\:=\:\frac{1200}{2401}\:\approx\:50\%\)