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I think I have seen this type of question before.
Say 8 persons are A,B,C,D,E,F,G,H.. AB,CD,EF,GH are couples
Calculate by reducing number of ways when couples will be together from number of total ways.
Number of total ways..8C3
Number of ways when couples will be together.. Say A, B are together..in this case number of ways to select 3rd person from remaining 6 is 6C1=6
Similarly when C,D are together..number of ways to selct 3rd person is 6.
When E, F are together..another 6 ways.
G,H are together..another 6 ways..
So total 24 ways.. when couples are together.
Hence required number of ways = 8C3 -24
So 32 should be the answer
Consider the couples as (H1,W1), (H2,W2), (H3,W3), (H4,W4), (H5,W5)
For each pair of couple they can pair up with either the husband or wife of another pair to form a group of 3
So for example, H1,W1,H2, H1,W1,W2..etc. So one couple can have total of 8 possible pairings
5 couples will give 5*8=40 pairs
So total number of 3 people group with husband and wife not in the saem group = 120-40 = 80
The undesirable outcome should have a couple. We can choose one out of 5 couples in 5C1 ways thus filling two spots(not one spot). Now, 10-2 = 8 members can be fill up the third spot in 8C1 ways. Thus total ways of undesired outcomes = 5C1*8C1 = 40.
Do let me know if I am not clear.
mallelac, do you mind explaining how you came to the formula for the undesireable outcomes? Wouldn't 5C1+8C1 only give you the number of people that would comprise a 2 person team?