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PS - Combinatorics (m02q05) [#permalink]
 12 Nov 2007, 18:36
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 Source: GMAT Club Tests - hardest GMAT questions
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walker wrote: C.
4C3*2C1*2C1*2C1=4*2^3=32
4C3 - 3 pairs from 4 pairs.
2C1 - changing between brother and sister in each pair. Right. same as mine. A(_ _) B(_ _) C(_ _) D(_ _) from A B C D, choose 3, 4C3 from the chosen 3, pick one from each 2C1*2C1*2C1 in total: 4C3*2C1*2C1*2C1=32
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Total possibilities excluding consideration for the restrictions :
8C3 = 8!/[3!x5!] = 8x7x6/3x2 = 56
The exluded possibilities due to the restriction (i.e. no sibiling in the 3-members committee ):
total pairs of sibilings = 4
corresponding possibilities for each pair (the 3rd member) = 6
So it is 4x6 = 24
ANSWER = total possibilities - restricted possibilities = 56 - 24 = 32
ANSWER: C
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C.
4C3*2C1*2C1*2C1=4*2^3=32
4C3 - 3 pairs from 4 pairs.
2C1 - changing between brother and sister in each pair.
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Re: PS - Combinatorics (m02q05) [#permalink]
28 Sep 2009, 04:22
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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
Soln: 8 * 6 * 4/3!
32 ways.
Ans is C
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Re: PS - Combinatorics (m02q05) [#permalink]
07 Jan 2010, 06:48
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yuefei wrote: 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 Source: GMAT Club Tests - hardest GMAT questions consider four siblings b1g1 b2g2 b3g3 b4g4 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.
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Re: PS - Combinatorics (m02q05) [#permalink]
07 Jan 2010, 10:21
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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.
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Answer 24
4c2 * 2c1 + 4c2* 2c1 (or 4* 3c2) = 24
Amar
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Answer C 32
The committee could have
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 )
Therefore 4+12+12+4=32 possible teams.
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pmenon wrote: 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.
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pmenon wrote: jbs wrote: pmenon wrote: 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.
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Re: PS - Combinatorics [#permalink]
12 Nov 2007, 20:46
yuefei wrote: 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
8C3 - (4C2 * 6C1)
56 - 4*6 = 56 - 24 = 32 (answer C)
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Re: PS - Combinatorics [#permalink]
13 Nov 2007, 10:46
beckee529 wrote: yuefei wrote: 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 8C3 - (4C2 * 6C1) 56 - 4*6 = 56 - 24 = 32 (answer C) beckee, why 4C2 and 6C1 ? I mean, can you break down how you came up with those #s ?
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Re: PS - Combinatorics [#permalink]
13 Nov 2007, 11:14
pmenon wrote: 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
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jimjohn wrote: can someone plz help explain what i did wrong?
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
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I understand that there are 4 pairs of siblings, but how can 4C2 = 4? Am I missing something here?
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Re: PS - Combinatorics [#permalink]
14 Nov 2007, 07:19
yuefei wrote: pmenon wrote: 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???
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Re: PS - Combinatorics [#permalink]
14 Nov 2007, 09:32
Skewed wrote: yuefei wrote: pmenon wrote: 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??? It should be 8C3 - 4C1*6C1
4C1 because you're picking one pair from 4 pairs.
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Mishari wrote: Total possibilities excluding consideration for the restrictions :
8C3 = 8!/[3!x5!] = 8x7x6/3x2 = 56
The exluded possibilities due to the restriction (i.e. no sibiling in the 3-members committee ):
6 possible people, we choose 1 of them.
6C1=6
corresponding possibilities for each pair (the 3rd member) = 6
we can do this with each pair of the twins.
total pairs of siblings = 4
(Unfavorable pairs)(6C1)
So it is 4x6 = 24
ANSWER = total possibilities - restricted possibilities = 56 - 24 = 32
ANSWER: C
excellent answer.
small elaboration above.
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Re: PS - Combinatorics [#permalink]
26 Aug 2008, 14:23
yuefei wrote: 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 = 8C1*6C1*4C1/3! = 32
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Re: PS - Combinatorics
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26 Aug 2008, 14:23
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