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0 boys and 3 girls = 4 possibilities ( 4c3 )
1 boy and 2 girls = 12 possibilities ( no siblings should be in the team : 4 * 3c2 )
2 boys and 1 girl = 12 possibilities ( no siblings should be in the team : 4c2 *2c1 )
3 boys and 0 girls = 4 possibilities ( 4c3 )

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

another approach.. someone please correct me if my logic is incorrect

total combinations - one pair of siblings

total = 8C3
1 pr of siblings = 4C2 * 6C1
4C2 = possible # of selecting a pair of siblings
6C1 = from the 6 left after pair is chosen, # of selecting the third

committee of 3 _ _ _ Number of ways of no siblings = total ways - ways in which at least 1 sibling features in the committee total ways=C(8,3)=56 At least 1= 4(any pair can be picked to be in the committee) * C(6,1) (one place to filled from 6 members)=24

answer C 32.
_________________

--------------------------------------------------------------------------------------- If you think you can,you can If you think you can't,you are right.

The third way is also valid, but you should divide 8C1x6C1x4C1 by 3! to get rid of duplications.

With 8C1x6C1x4C1 you can have ABC members as well as BCA members, which is basically the same group.

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

If you are collecting the methods to solve this problem here is another one:

4C3*2^3=32

4C3 - # of ways to select the sibling pair, which will be "granted" the right to give member; 2^3 - each selected sibling pair can give either brother or sister for membership 2*2*2=2^3.

The third way is also valid, but you should divide 8C1x6C1x4C1 by 3! to get rid of duplications.

With 8C1x6C1x4C1 you can have ABC members as well as BCA members, which is basically the same group.

Bunnel, how you came to the conclusion that 8C1x6C1x4C1 should be divided by 3! to get rid of duplications. Why not 4! or 2!. Please explain.

We divide by the number of members in the committee, so by 3!.

Consider this: 8C1*6C1*4C1 will give you all committees of ABC possible - (ABC), (ACB), (BAC), (BCA), (CAB) and (CBS) which are the SAME committee of 3 (3 distinct letters can be arranged in 3! ways). So we should divided 8C1*6C1*4C1 by 3!.

This is the way I was thinking about it, but can someone explain why it is wrong:

For the first member, there are 8 ways to select them.

For the second member, we can select from the remaining 7 people, but not the sibling of the first, so that leaves us with selecting from 6 people

For the final member, choose from remaining 5-1=4 people

Therefore 8x6x4

Let the brothers be A,B,C,D and the sisters be E,F,G,H.

With your method, you will have committees like
ABF, AFB, BAF,...... Basically, you will end up counting committees (that have the same members) multiple times.

This is the way I was thinking about it, but can someone explain why it is wrong:

For the first member, there are 8 ways to select them.

For the second member, we can select from the remaining 7 people, but not the sibling of the first, so that leaves us with selecting from 6 people

For the final member, choose from remaining 5-1=4 people

Therefore 8x6x4

Let the brothers be A,B,C,D and the sisters be E,F,G,H.

With your method, you will have committees like ABF, AFB, BAF,...... Basically, you will end up counting committees (that have the same members) multiple times.

Thanks, jbs. For this type of question, does AFB and ABF not count as a "different" committee ?

Also, if I were to go along my approach, how would I be able to remedy it to come up with the proper answer ?

They asked for the committee as a whole. So yes, for this type of question, AFB and ABF do not count as different committees. If they had said something w.r.t positioning , the problem would have changed.

Consider 3 members A,B & C.

Among themselves, A,B & C can form the following similar committees:

ABC
ACB
BAC
BCA
CAB
CBA

i.e: each group of 3 members can form 6 committees in which they repeat together but in different positions.

Therefore, in general, to correct your approach, divide your answer with the number of possible repetitions i.e. 6.

Hope this helps.

p.s: Your approach would have been correct in problems where positioning matters.

Last edited by jbs on 12 Nov 2007, 20:43, edited 1 time in total.

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

another approach.. someone please correct me if my logic is incorrect

total combinations - one pair of siblings

total = 8C3 1 pr of siblings = 4C2 * 6C1 4C2 = possible # of selecting a pair of siblings 6C1 = from the 6 left after pair is chosen, # of selecting the third

beckee, why 4C2 and 6C1 ? I mean, can you break down how you came up with those #s ?

4C2*6C1 is the total number of combinations with 2 siblings.

There are 3 people in each group.
4C2 is the number of sibling combinations for the first 2 spots in the group. Since we are looking for a pair instead of individual brothers & sisters, we are selecting from 4 instead of 8.
6C1 will fill the final spot (only 6 left here since we filled the first 2 spots with siblings).

there are eight possible choices for the first committee member

for the second committee member, we can't have the first person's sibling on the team so we have six possible choices for the second committee member

for the third committee member we can't have either of the first two chosen members's siblings, so we have four possible choices for the third committee member

so the total num of ways = 8*6*4

Ah ha ! I knew I couldnt have been the only person who thought of it like this !

Scroll up to view my posts, along with jbs'. We had a mini conversation about how to make this approach work, and why it wouldnt work as 8 x 6 x 4

beckee, why 4C2 and 6C1 ? I mean, can you break down how you came up with those #s ?

4C2*6C1 is the total number of combinations with 2 siblings.

There are 3 people in each group. 4C2 is the number of sibling combinations for the first 2 spots in the group. Since we are looking for a pair instead of individual brothers & sisters, we are selecting from 4 instead of 8. 6C1 will fill the final spot (only 6 left here since we filled the first 2 spots with siblings).

8C3 - 4C2*6C1 = 32

Isn't 8c3 - 4c2* 6c1 = 56 - 6*6 = 56-36= 24? How is the OA 32???

beckee, why 4C2 and 6C1 ? I mean, can you break down how you came up with those #s ?

4C2*6C1 is the total number of combinations with 2 siblings.

There are 3 people in each group. 4C2 is the number of sibling combinations for the first 2 spots in the group. Since we are looking for a pair instead of individual brothers & sisters, we are selecting from 4 instead of 8. 6C1 will fill the final spot (only 6 left here since we filled the first 2 spots with siblings).

8C3 - 4C2*6C1 = 32

Isn't 8c3 - 4c2* 6c1 = 56 - 6*6 = 56-36= 24? How is the OA 32???