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# M02-05

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Manager
Joined: 31 Oct 2016
Posts: 106

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15 Jun 2018, 00:34
3C8 - total number of ways to select committee of 3 people = 56
4! = 24 number of ways to select sibling pairs.
56-24 = 32
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Joined: 12 Nov 2016
Posts: 7

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10 Jun 2019, 21:43
Bunuel wrote:
Official Solution:

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

Since the committee shouldn't have siblings in it, then each pair can send only one "representative" to the committee. # of ways to choose which 3 pairs of brothers and sisters will send one "representative" to the committee is $$C^3_4$$ (choosing 3 pairs which will be granted the right to send one "representative" to the committee).

But each of these 3 pairs can send 2 persons to the committee either a brother or a sister: $$2*2*2=2^3$$.

So total # of ways to form the committee is $$C^3_4*2^3=32$$.

Hello Bunuel

Could you please look into below mentioned interpretation and let me know where I went wrong
Let there be 4 pairs of bros and sis : B1S1; B2S2; B3S3; B4S4
Ways to select 1st member : 8
ways to select 2nd member : 6 (only 6 members remaining after excluding the first selection and corresponding sibling)
Ways to select 3rd member : 4 (only 4 members remaining after excluding the first and second selection and their corresponding siblings)
Total ways = 8*6*4 = 192 ways
Math Expert
Joined: 02 Sep 2009
Posts: 56304

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10 Jun 2019, 23:39
1
1
pm0103 wrote:
Bunuel wrote:
Official Solution:

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

Since the committee shouldn't have siblings in it, then each pair can send only one "representative" to the committee. # of ways to choose which 3 pairs of brothers and sisters will send one "representative" to the committee is $$C^3_4$$ (choosing 3 pairs which will be granted the right to send one "representative" to the committee).

But each of these 3 pairs can send 2 persons to the committee either a brother or a sister: $$2*2*2=2^3$$.

So total # of ways to form the committee is $$C^3_4*2^3=32$$.

Hello Bunuel

Could you please look into below mentioned interpretation and let me know where I went wrong
Let there be 4 pairs of bros and sis : B1S1; B2S2; B3S3; B4S4
Ways to select 1st member : 8
ways to select 2nd member : 6 (only 6 members remaining after excluding the first selection and corresponding sibling)
Ways to select 3rd member : 4 (only 4 members remaining after excluding the first and second selection and their corresponding siblings)
Total ways = 8*6*4 = 192 ways

We need to divide $$8*6*4=192$$ by the factorial of the # of people - 3! to get rid of duplications $$8*6*4=192$$ contains ---> $$\frac{192}{3!}=32$$ - correct answer.

Consider example with smaller number: there are two couples and we want to choose 2 people not married to each other.
Couples: $$A_1$$, $$A_2$$ and $$B_1$$, $$B_2$$. Committees possible:

$$A_1,B_1$$;
$$A_1,B_2$$;
$$A_2,B_1$$;
$$A_2,B_2$$.

Only 4 such committees are possible.

If we do as proposed in the solution you posted:
The first person on the committee can be anyone of the 4.
The second person on the committee can be only one out of 2 (the first person with her or his sibling excluded).

So we'll get: 4*2=8, so more than 4, which means that 8 contains some duplications. And to get the right answer we should divide 8 by 2! --> 8/2!=4.

It's because if you pick A1 for the first pick and than pick B1 for the second you'll get the committee {A1, B1} but if you'll pick B1 for the first pick and then A1 you'll get the exact same committee {A1, B1} (dividing by the factorial of the # of people in committee you'll exclude this double countings).

Hope it helps.
_________________
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Joined: 12 Nov 2016
Posts: 7

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11 Jun 2019, 00:04
Bunuel wrote:
pm0103 wrote:
Bunuel wrote:
Official Solution:

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

Since the committee shouldn't have siblings in it, then each pair can send only one "representative" to the committee. # of ways to choose which 3 pairs of brothers and sisters will send one "representative" to the committee is $$C^3_4$$ (choosing 3 pairs which will be granted the right to send one "representative" to the committee).

But each of these 3 pairs can send 2 persons to the committee either a brother or a sister: $$2*2*2=2^3$$.

So total # of ways to form the committee is $$C^3_4*2^3=32$$.

Hello Bunuel

Could you please look into below mentioned interpretation and let me know where I went wrong
Let there be 4 pairs of bros and sis : B1S1; B2S2; B3S3; B4S4
Ways to select 1st member : 8
ways to select 2nd member : 6 (only 6 members remaining after excluding the first selection and corresponding sibling)
Ways to select 3rd member : 4 (only 4 members remaining after excluding the first and second selection and their corresponding siblings)
Total ways = 8*6*4 = 192 ways

We need to divide $$8*6*4=192$$ by the factorial of the # of people - 3! to get rid of duplications $$8*6*4=192$$ contains ---> $$\frac{192}{3!}=32$$ - correct answer.

Consider example with smaller number: there are two couples and we want to choose 2 people not married to each other.
Couples: $$A_1$$, $$A_2$$ and $$B_1$$, $$B_2$$. Committees possible:

$$A_1,B_1$$;
$$A_1,B_2$$;
$$A_2,B_1$$;
$$A_2,B_2$$.

Only 4 such committees are possible.

If we do as proposed in the solution you posted:
The first person on the committee can be anyone of the 4.
The second person on the committee can be only one out of 2 (the first person with her or his sibling excluded).

So we'll get: 4*2=8, so more than 4, which means that 8 contains some duplications. And to get the right answer we should divide 8 by 2! --> 8/2!=4.

It's because if you pick A1 for the first pick and than pick B1 for the second you'll get the committee {A1, B1} but if you'll pick B1 for the first pick and then A1 you'll get the exact same committee {A1, B1} (dividing by the factorial of the # of people in committee you'll exclude this double countings).

Hope it helps.

Thanks a ton
Senior Manager
Status: Gathering chakra
Joined: 05 Feb 2018
Posts: 369

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09 Jul 2019, 08:42
Bunuel 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

Notes:
-- We are selecting, not arranging groups/people, so we use combinations
-- There is no replacement, in fact "choose 1 sibling" requires us to remove the other possible sibling for the next step
-- The 4 pairs are different, but the people in them count as identical (each is a sibling rather than boy/girl). So we are choosing 1 out of 2 identical people to put into 3 out of 4 different groups.

Ways to choose three pairs = 4C3 = 4
Ways to choose 1 sibling out of 2 from each of the three pairs = 2C1 * 2C1 * 2C1
4*8 = 32
Manager
Joined: 29 May 2018
Posts: 67
Location: India
GMAT 1: 700 Q48 V38
GPA: 4

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13 Jul 2019, 05:41
Bunuel

Although I got this correct, I was uncomfortable with 4 distinct pairs of brothers and sisters. If the pair has to be distinct it can well be A,B; B,C; C,A; D,A. Please see that pairs are distinct here.
Math Expert
Joined: 02 Sep 2009
Posts: 56304

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13 Jul 2019, 06:04
vinayakvaish wrote:
Bunuel

Although I got this correct, I was uncomfortable with 4 distinct pairs of brothers and sisters. If the pair has to be distinct it can well be A,B; B,C; C,A; D,A. Please see that pairs are distinct here.

If A is B sibling and B is C's sibling, then isn't A a sibling of C?
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Joined: 29 May 2018
Posts: 67
Location: India
GMAT 1: 700 Q48 V38
GPA: 4

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13 Jul 2019, 06:07
Bunuel wrote:
vinayakvaish wrote:
Bunuel

Although I got this correct, I was uncomfortable with 4 distinct pairs of brothers and sisters. If the pair has to be distinct it can well be A,B; B,C; C,A; D,A. Please see that pairs are distinct here.

If A is B sibling and B is C's sibling, then isn't A a sibling of C?

Yes! please see that I have mentioned C,A as a distinct pair. The point of the matter is C,A and A,B are distinct pairs
Re: M02-05   [#permalink] 13 Jul 2019, 06:07

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# M02-05

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