Let's Suppose we have 4 Couples.

Call them

M1 F1
M2 F2
M3 F3
M4 F4

Now, as per question statement, apply FCP:

we have three Slots

____ ____ ____


How many choices do we have for Slot 1? 8 Choices because any of the person from a couple can be selected for 1st Slot

=> 8

Now move on to the 2nd Slot. How many choices do we have for the 2nd Slot? Let's say we have selected M1 for Slot 1. Now, F1 can't be selected for 2nd Slot because of the restriction in the question statement.

For 2nd Slot, total number of choices will be

8 6

Similary, for 3rd Slot, Let's say we have selected F2 for Slot 2. Now, M2 can't be selected for 3rd Slot because of the restriction in the question statement.

Since, M1 and F2 are already selected by us for slot 1 and 2 respectively, F1 and M2 can't be selected for third slot, and M1 and F2 are already on the committee. We are left with only four choices which are: M3 F3 M4 F4

So, for Slot 3

8 6 4

Now, divide the whole expression by 3! to remove the repeatitions.

=> 8*6*4/6
=> 32

Hence E!

Take any 3 out of these 4 couples in 4C3 ways,

Now, from each couple you've to choose 1 person to represent. i.e., 2C1 for each couple and we have to choose one person per couple out of the three couples we've chosen above,

Therefore, the total number of ways in which the committee can be formed is - 4C3*2C1*2C1*2C1= 4*2*2*2 = 32 ways

Answer is E

Thanks

Sincerely

Daksh Kumar