5 couples say a b c d e

2 2 2 2 2

Now if we apply the condition then we can select one from each in 5C3 ways ..so i think it shd be 5c3 * 2 = 20

Please correct me if i am wrong

1. Selecting the first member - There are 10 ways of selecting the first person out of a total of 10 people.
2. Selecting the second member - Now you are left with only 8 members to select from because the spouse of the already selected member is not eligible.
3. Selecting the last member - Again the spouse of the 2nd member is not eligible so you are left with only 6 people to choose from.

So the total number of ways the committee can be formed is:
10*8*6 = 480.

However, the three members within each group can be arranged in 3! ways (meaning there are 6 occurences of the same group in the total).

Hence the total number of distinct groups that you can select is 480/6 = 80.

Regards,
Samsonite

Number of ways of picking the first group = 5
Number of ways of picking a member from this group = 2

Number of ways of picking the second group = 4
Number of ways of picking a member from this group = 2

Number of ways of picking the last group = 3
Number of ways of picking a member from this group = 2

Total number of arrangements = 5*2*4*2*3*2 = 480
Number of redundant groups (because H1W2H3 is the same as W2H1H3) = 3! = 6.

Hence the number of ways of picking 3 members is 480/6 = 80.

Remember: Combination = Permutation/redundancy

Regards,
Samsonite

Neat explanation ...point noted ..

Thanks guys!