i would use a permutation logic here first

the first person to be selected on committee is selected amongst ten people, 10*
the second person will need to be someone not married with the first person, hence one less from the residual 9, (9-1)*
the third person will be two less from the residual 8, (8-2)*
and the last will be three less from the residual 7, (7-3)*

10*8*6*4=1620

then discount the permutation by 4! ways of selecting the group, 1620/4!=80
d

jayanti wrote:If a committee of 4 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

20
40
50
80
120

jayanti wrote:If a committee of 4 people is to be selected from among 5 married couples so that the committee does not include two people who are married to each other, how many such committees are possible?

20
40
50
80
120

Number of choices for the 1st person = 10.
Number of choices for the 2nd person = 8. (9 people left, but we can't choose the spouse of the first committee member chosen, leaving 9-1= 8 choices.)
Number of choices for the 3rd person = 6. (8 people left, but we can't choose the spouses of the first 2 committee members chosen, leaving 8-2 = 6 choices.)
Number of choices for the 4th person = 4. (7 people left, but we can't choose the spouses of the first 3 committee members chosen, leaving 7-3 = 4 choices.)

To combine the number of choices we have for each position on the committee, we multiply the numbers above:
10*8*6*4.

But the order in which the committee members are chosen doesn't matter: selecting ABCD will yield the same committee as selecting BDAC.
So that we don't overcount these duplicate arrangements, we must divide by the number of ways to arrange 4 elements (4!).

Thus:
Total number of possible committees = (10*8*6*4)/4! = 80.

The correct answer is D.