There are 5 married couples and a group of three is to be formed out of them; how many arrangements are there if a husband and wife may not be in the same group?
I don't understand the answer. In fact, I think it's wrong.
Another way to approach this problem would be to use basic counting principles.
10 Available people
3 Available positions _ _ _
In the first position, you can put any one of the 10 people. In the second position, you can put any one of 8 people (not 9 because the spouse of the person put in 1st position cannot be put in the second position). In the third position you can put any one of 6 people (2 people put in previous places and their spouses cannot be put in the third position)
Total number of arrangement = 10*8*6
But mind you, these are arrangements. We just need a group of 3 people. They don't need to be arranged in the group. So we divide these arrangements by 3! to get just the combination.
Answer = 10*8*6/3! = 80
(I do hope I have read the question properly this time petrifiedbutstanding!)
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