Question

A family has four daughters. Their home has three bedrooms for the girls. Two of the bedrooms are only big enough for one girl. The other bedroom will have two girls. How many ways are there to assign the girls to bedrooms?

Answer 1

Answer: 12 different ways.

Step-by-step explanation:

For a set of M elements, the number of combinations of N elements (N ≤ M) is given by:

$C(M, N) = \frac{M!}{(M - N)!*N!}$

In this case we know that there is a room that can hold two of the girls, then the number of combinations of the two girls that can be in that room is:

$C(4, 2) = \frac{4!}{2!*2!} = \frac{4*3}{2} = 6$

So these are the different combinations for the larger room.

So we assume that two girls already have a room, then we have two girls left and two rooms left.

For the first room, there are two possible girls that can be in, let's choose one of them.

For the last room, whe have only one girl left, so we have one option.

The total number of combinations will be equal to the product between the number of options for each event (where the events are assigning the girls to the rooms)

Then the number of different outcomes is:

C = 6*2*1 = 12

There are 12 different ways.