Well, 5 couples means 10 people. So need a group of 3 from 10, which becomes 10!/(3!*7!) = 120 total possible groups.

We are not interested in having partners in the group however, so we must remove these occurrences. Each couple could be paired with 8 other people, so for each of the 5 couples there are 8 unacceptable groups, or 40 unacceptable groups.

120 - 40 = 80

So there are 80 acceptable groups.

Please explain me why 10!/(3!*7!) ??

B/c you need to choose 3 people out of 10

Here is a simpler way to solve the "married couples" class of problems. Thanks to Ron Purewal.

5 married couples => 10 ppl.

The first slot can be filled in 10 ways.
Second slot can be filled in 8 ways (ie. exclude the person in slot 1 and his/her spouse)

third slot can be filled in 6 ways (exclude the person and their spouses in slots 1 and 2)

Total arrangements = 10*8*6

Now we are looking for combination of 3 ppl. The above reflects permutations of grp of 3 ppl. To get the combination, we divide by 3!
ie
10*8*6/3! = 80

Ht helps