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Permutation/Combination Question

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by focusgmat » Tue Oct 12, 2010 5:16 am
Source : GMATCLUB tests

If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

OA C



Siblings A1 A2
B1B2 C1 C2 D1D2

First slot can be filled by any sibling so 8C1
Then second slot can be filled by anybody else than the sibling of the person selected in slot one - 6c1
third slot in the same fashion - 4c1
I got 8*6*4 = 192 as the answer.

Please help me understand.
My solution gave answer E
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by fskilnik@GMATH » Tue Oct 12, 2010 5:25 am
Hi, focusgmat!

Common mistake... and easy to explain! ;)

First note that your answer is exactly 6 times the right one...and that 6 is 3!, that is, the number of permutations of 3 different objects.

Therefore after your reasoning ends, you should say... "Now I divide by 3! to get the answer!".

Explanation:

Say you began choosing B1 (viable), then you choose C1 (again viable) then you choose D2 (again ok).
This should be ONE of the answers, but with your reasoning, you would also count the following choice sequence:

C1 (viable), then B1 (also) then D2 (also), but this choice HAS ALREADY BEEN COUNTED!

In other words, for every choice you make, you put 5 "EXTRA" identical choices in the package, and you shouldn´t!

I will post below another solution, that avoids this "final correction" you must make to turn your solution right, ok?

Regards and good studies!
Fábio.
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by fskilnik@GMATH » Tue Oct 12, 2010 5:33 am
Answer: C

Let us see a solution based in COMBINATIONS associated with ARRANGEMENTS, then it will be unecessary to make "the correction" (divide by permutation), as I explained to "focusgmat" in the post above.

The solution below is "physically" natural, that why I recommend it to my students!

(Phase 1) From the four couples of siblings, let us choose 3 of them that will have a representative as part of the committee.

This may be done in (obvious) 4 ways (there are 4 ways of excluding one couple...), but if you prefer, just calculate C(4,3) = 4.

(Phase 2) Considering the 3 couples that will bring a representative (each one from each couple) "in front of you", just see that you have 2 options of choice for each couple, therefore you have 2*2*2 = 8 options of choice (really different options) for EACH choice made at Phase 1.

From the Multiplicative Principle, you have 4*8 = 32 choices!
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by GMATGuruNY » Tue Oct 12, 2010 6:11 am
focusgmat wrote:Source : GMATCLUB tests

If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

OA C

Siblings A1 A2
B1B2 C1 C2 D1D2

First slot can be filled by any sibling so 8C1
Then second slot can be filled by anybody else than the sibling of the person selected in slot one - 6c1
third slot in the same fashion - 4c1
I got 8*6*4 = 192 as the answer.

Please help me understand.
My solution gave answer E
Perfect reasoning, but you determined the number of ways to arrange 3 non-siblings. We need to determine the number of ways to combine them.

While ABC, CAB, and BCA are all different arrangements, they represent the same combination of 3 people. Your result is counting them as different combinations. In order not to overcount the duplicate combinations, we need to divide the number of possible arrangements by (the number of elements being chosen)!. Since in the problem above we're choosing 3 people, we need to divide 192 by 3!.

192/3! = 32.

So there are 192 ways to arrange 3 non-siblings, but only 32 ways to combine them.

Please note that the number of possible combinations will always be smaller than the number of possible arrangements. Most combination questions will include -- as a trap answer -- the number of possible arrangements, which will be a larger number. If the question is asking for the number of possible combinations, be skeptical of the largest answer choice.
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by bdevas01 » Thu Oct 14, 2010 8:35 am
GMATGuruNY wrote:
focusgmat wrote:Source : GMATCLUB tests

If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

OA C

Siblings A1 A2
B1B2 C1 C2 D1D2

First slot can be filled by any sibling so 8C1
Then second slot can be filled by anybody else than the sibling of the person selected in slot one - 6c1
third slot in the same fashion - 4c1
I got 8*6*4 = 192 as the answer.

Please help me understand.
My solution gave answer E
Perfect reasoning, but you determined the number of ways to arrange 3 non-siblings. We need to determine the number of ways to combine them.

While ABC, CAB, and BCA are all different arrangements, they represent the same combination of 3 people. Your result is counting them as different combinations. In order not to overcount the duplicate combinations, we need to divide the number of possible arrangements by (the number of elements being chosen)!. Since in the problem above we're choosing 3 people, we need to divide 192 by 3!.

192/3! = 32.

So there are 192 ways to arrange 3 non-siblings, but only 32 ways to combine them.

Please note that the number of possible combinations will always be smaller than the number of possible arrangements. Most combination questions will include -- as a trap answer -- the number of possible arrangements, which will be a larger number. If the question is asking for the number of possible combinations, be skeptical of the largest answer choice.

Hey Guru, I calculated it like this:

Total number of possible 3 person groups: (8!)/(3!5!) = 56
Groups WITH siblings: 4 (pairs of siblings) x 6 (for any one of the six remaining people) = 24

56 - 24 = 32

Is there anything wrong with this method? Do you foresee any scenarios in which I may get the wrong answer using this method?
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by GMATGuruNY » Thu Oct 14, 2010 10:33 am
bdevas01 wrote:
GMATGuruNY wrote:
focusgmat wrote:Source : GMATCLUB tests

If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?

8
24
32
56
192

OA C

Siblings A1 A2
B1B2 C1 C2 D1D2

First slot can be filled by any sibling so 8C1
Then second slot can be filled by anybody else than the sibling of the person selected in slot one - 6c1
third slot in the same fashion - 4c1
I got 8*6*4 = 192 as the answer.

Please help me understand.
My solution gave answer E
Perfect reasoning, but you determined the number of ways to arrange 3 non-siblings. We need to determine the number of ways to combine them.

While ABC, CAB, and BCA are all different arrangements, they represent the same combination of 3 people. Your result is counting them as different combinations. In order not to overcount the duplicate combinations, we need to divide the number of possible arrangements by (the number of elements being chosen)!. Since in the problem above we're choosing 3 people, we need to divide 192 by 3!.

192/3! = 32.

So there are 192 ways to arrange 3 non-siblings, but only 32 ways to combine them.

Please note that the number of possible combinations will always be smaller than the number of possible arrangements. Most combination questions will include -- as a trap answer -- the number of possible arrangements, which will be a larger number. If the question is asking for the number of possible combinations, be skeptical of the largest answer choice.

Hey Guru, I calculated it like this:

Total number of possible 3 person groups: (8!)/(3!5!) = 56
Groups WITH siblings: 4 (pairs of siblings) x 6 (for any one of the six remaining people) = 24

56 - 24 = 32

Is there anything wrong with this method? Do you foresee any scenarios in which I may get the wrong answer using this method?
Perfect reasoning! You used a strategy that I use often:

Good = Total - Bad

The total number of good combinations = the total number of possible combinations - the total number of bad combinations.

The question is which method will be easier and/or more efficient: to determine the total number of good combinations directly (as I did above) or to determine the total number of possible combinations and subtract the bad ones (as you did). Either method is fine, as long as we answer the question correctly and efficiently. :)
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