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gmatclub combinatorics

by bblast » Wed May 18, 2011 11:49 pm
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

C
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by kevincanspain » Thu May 19, 2011 12:03 am
You could reason that we need to choose 3 of the 8 people but exclude the groups consisting of two siblings and a third person:

8C3 - 4 x 6 = (8 x 7 x 6)/3 x 2 x 1 - 24 = 32

Alternatively, imagine for a moment that the order of choice mattered: We need 3 unrelated people 8 x x 4. Since the order of choice does not matter, we need to divide by 3!
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by dv2020 » Thu May 19, 2011 12:04 am
There are three SLOTs to be filled none of the siblings can be on the committee and here order doesn't matter between the three selected members so we will have to divide by n!

Hence,

First place can be taken by any of the 8 members, next taken by remaining 6(one selected member and sibling gone) & the third slot taken by (8-2 set of siblings gone now=4)


8X6X4/3!=8X4=32
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by bblast » Thu May 19, 2011 12:42 am
kevin i did the easy part

wrote "8C3- ", but then could not figure out what to subtract. Why subtract 24 from it ?


DV2020
i got the part 8*6*4 but why divide by 3! ? is this because its a combo nd not permutation ?
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by dv2020 » Thu May 19, 2011 1:04 am
The idea is to figure out whether order matters or not...generally we use combinations formula when order doesn't matter and permutations when order does matter...if you can figure out that the order doesn't matter dividing by 3! is the same as the combinations formula


For Eg if brothers are B1 B2 B3 B4 & sisters S1 S2 S3 S4
so let us select a committee
B1B2S3
the five other orders within these 3 are not counted separately here but all six orders count as one
B1S3B2
B2B1S3
B2S3B1
S3B1B2
S3B2B1

So the order doesn't matter. hence we divide by 3!

to be ultra clear about using this method i recommend take some time and watch Ron Purewal's combinatoric study hall...I was also always confused about which formula Permutation or combinations which one to use but i found the answers after going though this study hall
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by kevincanspain » Thu May 19, 2011 1:46 am
kevincanspain wrote:You could reason that we need to choose 3 of the 8 people but exclude the groups consisting of two siblings and a third person:

8C3 - 4 x 6 = (8 x 7 x 6)/3 x 2 x 1 - 24 = 32

Alternatively, imagine for a moment that the order of choice mattered: We need 3 unrelated people 8 x 6 x 4. Since the order of choice does not matter, we need to divide by 3!
You could also reason that 3 of the 4 sibling pairs will have a member on the committee, and that each of these 3 pairs selected can 'donate' either of its two members:

4C3 x 2 x 2 x 2

Note that 4C3 =4C1 = 4
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by GMATGuruNY » Thu May 19, 2011 2:56 am
bblast wrote: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

C
Good committees = Total possible committees - Bad committees.

Total possible committees:
Total number of combinations of 3 that can be formed from 8 choices = 8C3 = 56.

Bad committee = a sibling pair combined with one other person:
Number of sibling pairs = 4.
Once a sibling pair has been chosen from the 8 people, the number of people left that can combined with the sibling pair = 8-2 = 6.
Multiplying, we get 4*6 = 24 bad committees.

Thus, good committees = total - bad = 56-24 = 32.

The correct answer is C.
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