$3$ locations for $4$ people

Question

If a room was divided into $3$ parts and $4$ people were told to stand in whichever part they want, what are the chances of having all $4$ people choose the same part, $3$ people choosing the same part and $2$ people choosing the same part?

Answer 1

Hint:

You may give each part of the room a number: 1,2,3.

Now, each person chooses one of the numbers. That way you obtain $4$-digit numbers consisting of the digits $1,2,3$.

For example, all being in one part of the room corresponds to the numbers $1111,2222,3333$.

Similarly, you may handle the other cases.

Answer 2

Total number of ways the the 4 people can choose is $6!/(2!*4!)=15$, if the 3 parts of the room are not interchangeable. There are 3 ways for all 4 people to be in same part, 6 ways for exactly 3 people in same part, and 6 ways for exactly 2 people in same part. The probabilities are, respectively, $\frac{3}{15},\frac{6}{15},\frac{6}{15}$.