# Math Help - combination 2

1. ## combination 2

4 men and their wives , 4 bachelors and 4 spinsters are travelling in two 8-seater compartments of a train , one of which is a smoking compartment while the other is not . In how many ways can the party be split up if no wife is separated from her husband ?

2. I'm not sure about the method, but here goes:

Pair up each husband and wife as a "couple". So we have 4 couples, 4 bachelors and 4 spinsters.

Effectively we have 12 "seats"

Now, let's take the following cases.

1. All couples in one compartment .
No. of ways this can be achieved is 2

2. Three couples in one compartment,

No. of combinations = $\binom{2}{1}\binom{4}{3}\binom{8}{2}$ =224

3. Two couples in one compartment

No. of combinations = $\binom{2}{1}\binom{4}{2}\binom{8}{4}$=840

4. One couple in one compartment

No. of combinations = $\binom{2}{1}\binom{4}{1}\binom{8}{6}$=224

Total combinations = 1. +2. +3. + 4. = 450+840 = 1290?