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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, 08:53
Hi Guys,
I have reverse question here if somebody can help.
I found 3 ways of calculating the answer:
1. 4C3 (All Boys)+4C3 (All Girls)+4C2x2C1(2 Boys and 1 girl)+2C1x4C2(2 Girl and 1 Boy) 2. (Total ways of making the committe)-( ways for selecting one pair)x(Selecting final committe member ) i.e [8C3-4C1x6C1] 3. This is tricky and I don't know why this is wrong (Ways of Selecting First Committe Member)x( Ways of selecting 2nd memeber)x(Ways of selecting 3rd member) i.e [8C1x6C1x4C1]
I need help proving 3rd way of calculation is wrong. Plz help
Re: PS - Combinatorics (m02q05) [#permalink]
07 Jan 2010, 09:17
First member can chosen from any of 8 Second - from the 6 left (we exclude the sibling). The third from the remaining 4. Finally, don't forget about repetitions like abc, acb...l number of which is 3!=6. So, (8*6*4)/6=32. Ans. is C.
Re: PS - Combinatorics (m02q05) [#permalink]
07 Jan 2010, 09:21
1
This post received KUDOS
Expert's post
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.
Re: PS - Combinatorics (m02q05) [#permalink]
12 Jan 2010, 07:14
I think that 4C2 should be 4C1 because we are only constraining the second seat on the committee from being a sibling before considering the third seat. 4C1*6C1 = 24.
Re: PS - Combinatorics (m02q05) [#permalink]
12 Jan 2011, 09:32
The First person can be chosen in 8 ways, the second person in 6 ways (the first persons sibling cannot be selected) and the third person in 4 ways
Total no of ways = 8*6*4 = 192. These include all the arrangement but we need only combination. Hence to get the no of combination divide the no of ways by 3!
Re: PS - Combinatorics (m02q05) [#permalink]
04 Jun 2011, 17:17
This is how I did it -
Consider 4 siblings: B1G1 B2G2 B3G3 B4G4
From any one pair of siblings, choose 1 person = 2C1 (From a brother & sister pair choose 1) We are selecting 3 people, so repeat this 3 times from 3 different pairs of siblings. So we get = 2C1 * 2C1 * 2C1 = 8
Now we have 4 pairs and we need to select any 3 pairs (From which to select 3 individual people - this part has been done in the above step) = 4C3 = 4
Re: PS - Combinatorics (m02q05) [#permalink]
04 Oct 2011, 19:56
Bunuel wrote:
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. _________________
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Re: PS - Combinatorics (m02q05) [#permalink]
20 Oct 2011, 17:14
Let the 4 groups be A B C D. Following are the 4 ways to select 3 people ABC, BCD, ACD,ABD. Since there are 2 ways to select a person from a group we have 8 different possibilities for each pairing. Since there are 4 pairings 8 x4=32 the final answer
Re: PS - Combinatorics (m02q05) [#permalink]
17 Jan 2012, 23:54
8*6*4/3!=32
u have to divide (8*6*4) by 3!, since all there are double counted. I mean first u chose any of 8, then any of 6. that any of 8 may consist any of 6. same with any of 4. _________________
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Re: PS - Combinatorics (m02q05) [#permalink]
18 Jan 2012, 03:30
Expert's post
GMATmission wrote:
Bunuel wrote:
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!.
Re: PS - Combinatorics (m02q05) [#permalink]
02 Feb 2012, 04:33
Sanjay76 wrote:
The First person can be chosen in 8 ways, the second person in 6 ways (the first persons sibling cannot be selected) and the third person in 4 ways
Total no of ways = 8*6*4 = 192. These include all the arrangement but we need only combination. Hence to get the no of combination divide the no of ways by 3!
The committee can be formed in = 192/6 = 32 ways
Sanjay
Thanks Sanjay, this was the easiest post to comprehend and helped me understand why they 3! was needed
2 ways: 1) Direct Choose 3 pairs out of 4 pairs = 4C3=4 Choose 1 person from each of 3 pairs = (2C1)^3 = 8 => no of committees as required = 4*8=32 2) Find the reverse Choose 3 ppl from 8 ppl = 8C3 = 56 Choose 3 ppl in with 1 pair of siblings = (4C1)*(6C1)=24 => 56-24=32