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# PS - Combinatorics (m02q05)

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Re: PS - Combinatorics (m02q05) [#permalink]

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28 Sep 2009, 03:22
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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]

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07 Jan 2010, 05:48
1
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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

[Reveal] Spoiler: OA
C

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

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Re: PS - Combinatorics (m02q05) [#permalink]

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07 Jan 2010, 06:38
1
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1. All possbilities
C3/8= 8!/3!*(8-3)!=56
2. eliminate the pairs with siblings:

AaB
AaC
Aab
Aac
...24 pairs
3. 56-24=32

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Re: PS - Combinatorics (m02q05) [#permalink]

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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

Thanks,
RRH
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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.
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07 Jan 2010, 09:21
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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.

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Re: PS - Combinatorics (m02q05) [#permalink]

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09 Jan 2010, 20:37
Mishari,

Total possibilities excluding consideration for the restrictions :
8C3 = 8!/[3!x5!] = 8x7x6/3x2 = 56

How come [3!x5!] yields 3x2?
Isnt it suppose to be 3!(1・2・3)×5!(1・2・3・4・5)??
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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.
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Re: PS - Combinatorics (m02q05) [#permalink]

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12 Jan 2011, 07:33
OA C = 32

Total Combinations - Not accepted
(8!/(5!3!)) - 4 x (6!/(5!1!))

Supose you pick two siblings, there are 6 candidates for the last post. This can occur four times as there are four pairs of brothers.
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Re: PS - Combinatorics (m02q05) [#permalink]

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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!

The committee can be formed in = 192/6 = 32 ways

Sanjay
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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

So we now get 8*4 = 32.
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Re: PS - Combinatorics (m02q05) [#permalink]

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04 Jun 2011, 22:36
4C1 * 3C2 + 4C1 * 3C2 + 4C3 + 4C3 = 32
1B, 2G
1G, 2B
3B
3G
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25 Sep 2011, 00:53
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 ):

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

How do you get corresponding possibilities for each pair (the 3rd member) = 6 ??? I dont understand
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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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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
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Re: PS - Combinatorics (m02q05) [#permalink]

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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]

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18 Jan 2012, 03:30
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!.

This question and the same doubt you have is also discussed here: if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html

Similar questions to practice:
ps-combinations-94068.html
ps-combinations-101784.html
committee-of-88772.html
if-4-people-are-selected-from-a-group-of-6-married-couples-99055.html
combination-permutation-problem-couples-98533.html

Hope it helps.
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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
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16 Jan 2013, 05:11
The three pair of sibling can be selected in 4C3 ways.
And among the three siblings, you can select one from each in 2 ways = 4C3 * 2 * 2 * 2 = 32
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Re: PS - Combinatorics (m02q05) [#permalink]

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20 Apr 2014, 05:32
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
Re: PS - Combinatorics (m02q05)   [#permalink] 20 Apr 2014, 05:32

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# PS - Combinatorics (m02q05)

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