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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 \(C^3_4*2^3=32\).

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)

I understand that you want to understand the total number of combinations of choosing 3 people from 4 pairs (so 4C3) and then we multiply that by 2^3 because for each of the three pairs they could either send the brother or sister - thus there's two combinations? Just want to make sure I'm thinking about this correctly.

Can you please suggest if the following approach is correct?

There are 4 sibling pairs

(B1,S1) (B2,S2) (B3,S3) (B4,S4)

Total Number of candidates - 8.

For filling the first slot in the committee, we have 8 people eligible. For filling the second slot in the committee, we have 6 people eligible. [Removing one of the sibling of the earlier selected candidate] For filling the third slot in the committee, we have 4 people eligible. [Removing another of the sibling of the earlier selected candidate]

Therefore total possibilities - 8*6*4.

However my answer does not figure in the options. Where am I going wrong?

Vijay

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 \(C^3_4*2^3=32\).

Can you please suggest if the following approach is correct?

There are 4 sibling pairs

(B1,S1) (B2,S2) (B3,S3) (B4,S4)

Total Number of candidates - 8.

For filling the first slot in the committee, we have 8 people eligible. For filling the second slot in the committee, we have 6 people eligible. [Removing one of the sibling of the earlier selected candidate] For filling the third slot in the committee, we have 4 people eligible. [Removing another of the sibling of the earlier selected candidate]

Therefore total possibilities - 8*6*4.

However my answer does not figure in the options. Where am I going wrong?

Vijay

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 \(C^3_4*2^3=32\).

Can you please suggest if the following approach is correct?

There are 4 sibling pairs

(B1,S1) (B2,S2) (B3,S3) (B4,S4)

Total Number of candidates - 8.

For filling the first slot in the committee, we have 8 people eligible. For filling the second slot in the committee, we have 6 people eligible. [Removing one of the sibling of the earlier selected candidate] For filling the third slot in the committee, we have 4 people eligible. [Removing another of the sibling of the earlier selected candidate]

Therefore total possibilities - 8*6*4.

However my answer does not figure in the options. Where am I going wrong?

Vijay

Vijay, Your approach is correct not complete . Since this is a selection problem, order of selection is not important. In your approach , you may have a committee with B1,S2,B3 as well as S2,B1,B3 .. if you notice this is due to the number of ways each is permitted. To eliminate this , you need to divide the result by 3!. Hope this helps.

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.

I found the following solution, bit lengthy but worked for me Four pairs means - 4B and 4S we can select 1B out of 4B -> 4C1 corresponding to that I have to select 2S out of 3S (because i have to exclude 1S corresponding to 1B, which are making pair) -> 3C2 So for the first combination it would be 4C1 * 3C2

B,S -> (1,2); (2,1); (3,0); (0,3)

The rest of the pairs would be (4C2*2C1); (4C3*1); (1*4C3)

So total required combination is 4C1*3C2 + 4C2*2C1 + 4C3*1 + 1*4C3 = 12+12+4+4=32

(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

I went through this problem in CAT: I broke the question in this way: 3 possibilities: 3 man or 3 girls or (2 man and 1 girl) or (2 girls and 1 man) Since or means add and we have 2 things that are ''simmetric'' combo of 3 man=combo of 3 girls=C4,3=4 combo of (2 man and 1 girl)=combo of (2 girls and a man) here the calculation for me was not immediate i did it this way: the combo of 2 man are C4,, now for each combo of man you can have 2 possible girls(the ones that are not sisters) so all the combo of 2 men and 1 girl( that as said are equal to the combo of 2 girls and one man) are:2C4,2 suming all up: 2*C4,3+2*2*C4,2=32

Easiest way to approach this is restriction method - Total number of combination that can be formed from 8 people - 8C3 Number of groups possible which includes atleast a brother -sister pair and a person from remaining - 4C1 ( choosing 1 out of 4 pairs of brother & sister) * 6C1 Final answer - 8C3 - (4C1*6C1) = 32

Please help understand where my thinking is flawed ? A 3 member committee needs to be formed from 4 pair of distinct siblings say (b1,s1), (b2, s2), (b3, s3), (b4,s4).

1st member can be any of the 8 persons. Now, for next member selection, discard the pair from which the 1st member was selected. So now we have 3 pairs left to choose from. 2nd member can be any of the 6 persons in the left 3 pairs. Now, for next member selection, discard the pair from which the 1st and 2nd member were selected. So now we have 2 pairs left to choose from. 3rd member can be any of the 4 persons in the remaining 2 pairs.

So total number of ways = 8*6*4 = 192. Where am I counting the extras ?

Please help understand where my thinking is flawed ? A 3 member committee needs to be formed from 4 pair of distinct siblings say (b1,s1), (b2, s2), (b3, s3), (b4,s4).

1st member can be any of the 8 persons. Now, for next member selection, discard the pair from which the 1st member was selected. So now we have 3 pairs left to choose from. 2nd member can be any of the 6 persons in the left 3 pairs. Now, for next member selection, discard the pair from which the 1st and 2nd member were selected. So now we have 2 pairs left to choose from. 3rd member can be any of the 4 persons in the remaining 2 pairs.

So total number of ways = 8*6*4 = 192. Where am I counting the extras ?