If a committee of 3 people is to be selected from among 5 married coup

In your selection of 5 people, say you selected 4 husbands and 1 wife (H1, H2, H3, H4, W5)
Now you selected 3 people out of these 5 say (H1, H2, H3)
This is one selection.

In your selection of 5 people, say you selected 3 husbands and 2 wives (H1, H2, H3, W4, W5)
Now you selected 3 people out of these 5 say (H1, H2, H3)
This is another selection.

But they both are the same selection but you are double counting them. That is why you are getting more than actual number.

I understood 3 approaches so far:
Approach #1:
Total # ways of forming committee of 3 members(without constraint) - # of ways in which 1 couple is present in the committee(constraint)
Total # ways of forming committee of 3 members(without constraint) = 10C3
# of ways in which 1 couple is present in the committee(constraint) = # of ways to select 1 couple out of 5 * # of ways of selecting 1 remaining individual out of 4 couples = 5C1 * 8C1

Approach #2:
# ways to select 3 couples out of 5 * # ways to select 1 person in C1 * ..C2 * ..C3 = 5C3*2C1*2C1*2C1

Approach #3:
We need to fill 3 seats in the committee
1st seat can be filled in 10C1 ways(selecting 1 out of 10 people)
2nd seat can be filled in 8C1 ways(selecting 1 out of 8 (=9 - spouse of 1st committee member))
3rd seat can be filled in 6C1 ways(selecting 1 out of 6 (=7 - spouse of 2nd committee member))
Total #of ways = 10C1*8C1*6C1
Now this answer includes order of selection as well, but we are not concerned about the order. # of ways to arrange 3 committee members is 3!. So dividing with the same to get the final answer

There are two distinct ways of approaching it:



­