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.
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