pm0103 wrote:

Bunuel wrote:

Official Solution:

There are four distinct pairs of brothers and sisters. In how many ways can a committee of 3 be formed and NOT have siblings in it?

A. 8

B. 24

C. 32

D. 56

E. 80

Since the committee shouldn't have siblings in it, then each pair can send only one "representative" to the committee. # of ways to choose which 3 pairs of brothers and sisters will send one "representative" to the committee is \(C^3_4\) (choosing 3 pairs which will be granted the right to send one "representative" to the committee).

But each of these 3 pairs can send 2 persons to the committee either a brother or a sister: \(2*2*2=2^3\).

So total # of ways to form the committee is \(C^3_4*2^3=32\).

Answer: C

Hello Bunuel

Could you please look into below mentioned interpretation and let me know where I went wrong

Let there be 4 pairs of bros and sis : B1S1; B2S2; B3S3; B4S4

Ways to select 1st member : 8

ways to select 2nd member : 6 (only 6 members remaining after excluding the first selection and corresponding sibling)

Ways to select 3rd member : 4 (only 4 members remaining after excluding the first and second selection and their corresponding siblings)

Total ways = 8*6*4 = 192 ways

We need to divide \(8*6*4=192\) by the factorial of the # of people - 3! to get rid of duplications \(8*6*4=192\) contains ---> \(\frac{192}{3!}=32\) - correct answer.

Consider example with smaller number: there are two couples and we want to choose 2 people not married to each other.

Couples: \(A_1\), \(A_2\) and \(B_1\), \(B_2\). Committees possible:

\(A_1,B_1\);

\(A_1,B_2\);

\(A_2,B_1\);

\(A_2,B_2\).

Only 4 such committees are possible.

If we do as proposed in the solution you posted:

The first person on the committee can be anyone of the 4.

The second person on the committee can be only one out of 2 (the first person with her or his sibling excluded).

So we'll get: 4*2=8, so more than 4, which means that 8 contains some duplications. And to get the right answer we should divide 8 by 2! --> 8/2!=4.

It's because if you pick A1 for the first pick and than pick B1 for the second you'll get the committee {A1, B1} but if you'll pick B1 for the first pick and then A1 you'll get the exact same committee {A1, B1} (dividing by the factorial of the # of people in committee you'll exclude this double countings).

Hope it helps.