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Re: PS - Combinatorics (m02q05) [#permalink]
23 Aug 2009, 11:34

xALIx wrote:

Well put!!

bmwhype2 wrote:

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.

I like this explanation but i am hung up on "6 possible people, we choose 1 of them"

if there are 8 people total and you want to form 3 person committees from 4 pairs of siblings who cannot be together in a committee and if you have you have 6 possible people (i'm assuming you've picked 1 person and the person's sibling is excluded) shouldn't we be choosing 2 people out of the 6 people that are left to form the 3 person committee?

Please go really slowly, i am far from a math whiz.

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]
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. _________________

Ifmypostdida dancein your mind, send methe stepsthrough kudos :)

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. _________________

Happy are those who dream dreams and are ready to pay the price to make them come true

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