Permutation and combination

This is a good one!

Rather than getting into formulas, let's think about it logically:

For the first person, we have 8 choices (we can pick anyone).

For the second person, we have 6 choices (we can pick anyone except the person already picked and his/her spouse).

For the third person, we have 4 choices (we can pick anyone except the two people already picked and their two spouses).

Thus, we have 192 different arrangements. But, since "order doesn't matter" in this case (e.g., a committee with Al, Bob, and Carl is the same as a committee with Carl, Bob, and Al), we have to divide.

Since we have three "items" in our calculation (the three committee members), we must divide 192 by 3! = 32.

priyank.hirani wrote:A committee of three people is to be chosen from 4 married couples. What is the
number of different committees that can be chosen if two people who are married
to each other cannot both serve on the committee?

16, 24, 26, 30, 32

Choosing I person...It can be anyone among 8 people = 8C1....(1);
Choosing II person...It can be anyone among 6 people = 6C1....(2);(As the spouse of selected to be lest)
Choosing III person...It can be anyone among 8 people = 4C1....(3);(As the spouse of selected to be lest)

Choosing 3 persons = 8C1*6C1*4C1 =192 ways.

Another approach to solve this question is -

total no of ways of selecting 3 people from 8 = 8C3 = 56

Now if the siblings are present in the combination then 2 places are booked and only one place is left to be filled....

1*1*6 for each sibling pair..We have 4 pairs ...so total number of combinations in which sibling pair will be present is -

4*1*1*6 = 24

Hence the number of combinations in which the sibling pair will not be presnet = 56 - 24 = 32

Hope this helps!

Shalabh's Quants wrote:

priyank.hirani wrote:A committee of three people is to be chosen from 4 married couples. What is the
number of different committees that can be chosen if two people who are married
to each other cannot both serve on the committee?

16, 24, 26, 30, 32

Choosing I person...It can be anyone among 8 people = 8C1....(1);
Choosing II person...It can be anyone among 6 people = 6C1....(2);(As the spouse of selected to be lest)
Choosing III person...It can be anyone among 8 people = 4C1....(3);(As the spouse of selected to be lest)

Choosing 3 persons = 8C1*6C1*4C1 =192 ways.

This would be the total number of arrangements.
Dont forget to divide it by 3! to get the number of selections.