Total ways=8c3=56 ways
now, let us take ways in which a couple is always present in committee, for this let us take couples as one object
i.e. Ways when couple present in committee = 4c1*6c1 =24
Ways when no couple present = 56-24 =32
IMO E=32

Yes, the answer is E.

Can u explain a little more please on the 4c1*6c1?

now, let us take ways in which a couple is always present in committee, for this let us take couples as one object
i.e. Ways when couple present in committee = 4c1*6c1 =24

cramya wrote:Can u explain a little more please on the 4c1*6c1?

now, let us take ways in which a couple is always present in committee, for this let us take couples as one object
i.e. Ways when couple present in committee = 4c1*6c1 =24

Hi cramya, sorry for delayed reply.
4c1*6c1 .....
the 4c1 parts comes from considering couples as one object i.e instead of 8 people we consider 4 people only... ways to selects 1 out of 4 is 4c1 (this will give us couple in committee of 3
we are left with 3rd place in the committe to be filled by any remaining 6 people = 6c1
therefore , total ways when atleast one couple is present =4c1*6c1 =24
hope its clear now

Thanks Earth@work!