Number of possibilities

To families are chosen at random. Each family has one male and one female.

Given that only one male out of both families and one female from both families have jobs, find the number of different arrangements for those with jobs.

-------
OK

So the arrangement is

(HW') (H'W)
OR
(H'W) (HW')
OR
(H'W') (HW)
OR
(HW) (H'W')

so there are four options

Is there a better way to work this out other than trial and error?

Reply 1

ghostwalker

The fact that they are in families is, in effect, irrelevant to the question.

There are 2 males and 2 females. One from each works, hence:

^2C_1\times ^2C_1

Reply 2

username459260

can you please explain whY???

Reply 3

ghostwalker

Original post by jsmith6131

can you please explain whY???

Which bit?

why it is 2C1 * 2C1

Original post by jsmith6131

why it is 2C1 * 2C1

these are the different way in which you can pick 1 from 2, since there are 2 couples, you multiply one by the other. You have 4 different combinations. 1