M02-05

Since we have to select 3 with NOT a sibling pair, then:

Total no of possible ways - number of ways of selecting 3 including a sibling pair

8C3 - 4C1 * 6C1

(No. of ways of selecting 8 taking 3 at a time) - Number of ways of selecting a pair out of 4, multiplied with selecting the remaining one person out of the left 3 pairs
(6 people)

Therefore: 56 - (4*6) = 56-24 = 32

Total # of committees = \(^8C_3\) = 56
Committees with 1 sibling = \(^4C_1 * ^6C_1\) = 24

56 -24 = 32

Answer C

Is this a possible correct solution as well ?

8*6*4
-------
3!

Thanks you

I attacked this question in a different way:
we have 8 people, out of which we have to choose 3. Thus, we have 8C3 ways to do so:
8C3 = 8*7*6/3! = 56.
Note that there are way fewer ways to choose in such ways that no brothers are selected.

We can eliminate D and E.

now, look at B and C. B+C = 56, clearly, the 2 options are complementary, and there is a high chance that the answer is one of the two. A is thus out.
It's a good way to make a strategic guess, chances to answer correctly - 50%

now, in how many ways can we select 3 people so that no brothers/sisters are in the committee?
well, let's see how many committees, in which the restriction is not respected, we can get!

now, there are 4 brothers and 4 sisters!
suppose we select 1 brother and 1 sister -> we then have 1*4C1*6C1 or 4*6= 24 ways, we're almost there!
why 4C1 - from 4, we have to choose 1 that will be brother/sister of the other one.
now, we have chosen 2 members, out of 6, we have to choose 1. that's why 6C1
alternatively, we can assign letters and try to list all the combinations, I know it is not preferable, but still 24 - not that much!

A1B1
A2B2
A3B3
A4B4

now, we can have A1B1*6C1; A2B2*6C1; A3B3*6C1; A4B4*6C1 -> clearly we have 4*6C1!!! It doesn't matter the order A1B1 ir B1A1:)



now, the final step:
56 ways total - 24 ways in which the restriction is not respected = 32 ways in which restriction is respected.

Another solution:

1st person: There are 8 ways of selection.

2nd person: Remaining 7 people, but not choosing the person who is pair matching with 1st, so there are 6 ways of selection.

3rd person: Remaining 6 people, not choosing the people who are pair maching with 1st and 2nd, so there are 4 ways of selection.

In addition, choosing no order, divided 3!

The no. of ways = 8*6*4/3! = 32 ways.

(B1,S1), (B2,S2), (B3,S3), (B4,S4) = 04 PAIR, TOTAL 08 PERSON

so total combination of 03 person 8c3= 56
LETS SAY NOW (B1,S1) one pair, so we can choose the third place from 06 person now, means when first borther and sisiter are always the part of the combination- third spot can be choosen by any of the remaining 06

ii) if (b2, s2) pair is choosen - so rest from 06, so total 06 combination

iii) (b3,s3) = 06

(iv) (b4,s4)= 06

so total combination when pair is always there = 06*04= 24

so total without sibling = 56-24= 32

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

Notes:
-- We are selecting, not arranging groups/people, so we use combinations
-- There is no replacement, in fact "choose 1 sibling" requires us to remove the other possible sibling for the next step
-- The 4 pairs are different, but the people in them count as identical (each is a sibling rather than boy/girl). So we are choosing 1 out of 2 identical people to put into 3 out of 4 different groups.

Ways to choose three pairs = 4C3 = 4
Ways to choose 1 sibling out of 2 from each of the three pairs = 2C1 * 2C1 * 2C1
4*8 = 32

We need 3 people from those 8. But we have the restriction to choose only one of each pair, so we have to choose 3 pairs for our selection.
Number of ways to choose 3 pairs: 4C3

For every case where we selected 3 pairs, we now have to select one out of every pair, since this is the only way we satisfy the given constraint.
For every of those 3 pairs, select one out of two. 3 * 2C1.

Combine: 4C3 * 2C1 * 3 = 32

I think this is a high-quality question and I agree with explanation.

A1B1
A2B2
A3B3
A4B4

Select 3 people without any sibling in it:
so i choose
8 * 6* 4= 192

why do i need to divide by 3!
Another approach:
Select 3 pairs
4*3*2 and in each pair it can be brother or sister
so i multiple by 2*2*2
again 192

i know i am wrong, please suggest what logic i am missing?
Please suggest chetan2u, Bunuel

When you take 8*6*4, order matters, but a committee has all members equal, so we look for combinations as order does not matter.
These 3 people can be arranged in 3! ways and that’s why we divide by 3!

The 192 ways takes A1,B2,C1 and B2,A1,C1 and A1,C1,B2 as separate cases.

Diff arrangements

1.2B+1G=4C2*2C1=12(If 2 boys are selected from 4,we should leave their respective siblings when choosing for the girl)
2.1B+2G=4C1*3C2=12(applying same logic as in 1)
3.3B+0G=4C3=4
4.0B+3G=4C3=4

Hence all possible groupings 12+12+4+4=32

Hope this helps!!

Posted from my mobile device

I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.

I think this is a high-quality question and I agree with explanation.

There are 4 sibling pairs. Let us represent them in the following way!

B1S1 B2S2 B3S3 B4S4

Now we have to choose 3 persons from these groups such that no one of them belong to a sibling pair.
Ways:
All 3 brothers: (4C3) = 4 [Choosing 3 brothers from 4]
All 3 sisters: (4C3) = 4 [ Choosing 3 sisters from 4]
2 brothers and 1 sister: (4C2)*(2C1) = 6*2 = 12 [ the 2 pairs from which the brothers have been chosen, sisters don't have those options to belong to, so they have remaining 2 pairs]
1 brother and 2 sisters: (4C1)*(3C2) = 4*3 = 12 [ the 1 pair from which the brother have been chosen, sisters don't have that options to belong to, so they have remaining 3 pairs]

So total possibilities = 4+4+12+12 = 32

I think this is a high-quality question and I agree with explanation.