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PS - Combinatorics (m02q05)

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PS - Combinatorics (m02q05) [#permalink] New post 12 Nov 2007, 17:36
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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

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 [#permalink] New post 12 Nov 2007, 17:45
Answer 24

4c2 * 2c1 + 4c2* 2c1 (or 4* 3c2) = 24

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 [#permalink] New post 12 Nov 2007, 17:57
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Answer C 32

The committee could have

0 boys and 3 girls = 4 possibilities ( 4c3 )
1 boy and 2 girls = 12 possibilities ( no siblings should be in the team : 4 * 3c2 )
2 boys and 1 girl = 12 possibilities ( no siblings should be in the team : 4c2 *2c1 )
3 boys and 0 girls = 4 possibilities ( 4c3 )

Therefore 4+12+12+4=32 possible teams.
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 [#permalink] New post 12 Nov 2007, 19:20
pmenon wrote:
This is the way I was thinking about it, but can someone explain why it is wrong:

For the first member, there are 8 ways to select them.

For the second member, we can select from the remaining 7 people, but not the sibling of the first, so that leaves us with selecting from 6 people

For the final member, choose from remaining 5-1=4 people

Therefore 8x6x4


Let the brothers be A,B,C,D and the sisters be E,F,G,H.

With your method, you will have committees like
ABF, AFB, BAF,...... Basically, you will end up counting committees (that have the same members) multiple times.
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 [#permalink] New post 12 Nov 2007, 19:39
pmenon wrote:
jbs wrote:
pmenon wrote:
This is the way I was thinking about it, but can someone explain why it is wrong:

For the first member, there are 8 ways to select them.

For the second member, we can select from the remaining 7 people, but not the sibling of the first, so that leaves us with selecting from 6 people

For the final member, choose from remaining 5-1=4 people

Therefore 8x6x4


Let the brothers be A,B,C,D and the sisters be E,F,G,H.

With your method, you will have committees like
ABF, AFB, BAF,...... Basically, you will end up counting committees (that have the same members) multiple times.


Thanks, jbs. For this type of question, does AFB and ABF not count as a "different" committee ?

Also, if I were to go along my approach, how would I be able to remedy it to come up with the proper answer ?


They asked for the committee as a whole. So yes, for this type of question, AFB and ABF do not count as different committees. If they had said something w.r.t positioning , the problem would have changed.

Consider 3 members A,B & C.

Among themselves, A,B & C can form the following similar committees:

ABC
ACB
BAC
BCA
CAB
CBA

i.e: each group of 3 members can form 6 committees in which they repeat together but in different positions.

Therefore, in general, to correct your approach, divide your answer with the number of possible repetitions i.e. 6.

Hope this helps.

p.s: Your approach would have been correct in problems where positioning matters.

Last edited by jbs on 12 Nov 2007, 19:43, edited 1 time in total.
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Re: PS - Combinatorics [#permalink] New post 12 Nov 2007, 19:46
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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


another approach.. someone please correct me if my logic is incorrect

total combinations - one pair of siblings

total = 8C3
1 pr of siblings = 4C2 * 6C1
4C2 = possible # of selecting a pair of siblings
6C1 = from the 6 left after pair is chosen, # of selecting the third

8C3 - (4C2 * 6C1)
56 - 4*6 = 56 - 24 = 32 (answer C)
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Re: PS - Combinatorics [#permalink] New post 13 Nov 2007, 09:46
beckee529 wrote:
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


another approach.. someone please correct me if my logic is incorrect

total combinations - one pair of siblings

total = 8C3
1 pr of siblings = 4C2 * 6C1
4C2 = possible # of selecting a pair of siblings
6C1 = from the 6 left after pair is chosen, # of selecting the third

8C3 - (4C2 * 6C1)
56 - 4*6 = 56 - 24 = 32 (answer C)


beckee, why 4C2 and 6C1 ? I mean, can you break down how you came up with those #s ?
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Re: PS - Combinatorics [#permalink] New post 13 Nov 2007, 10:14
pmenon wrote:
beckee, why 4C2 and 6C1 ? I mean, can you break down how you came up with those #s ?


4C2*6C1 is the total number of combinations with 2 siblings.

There are 3 people in each group.
4C2 is the number of sibling combinations for the first 2 spots in the group. Since we are looking for a pair instead of individual brothers & sisters, we are selecting from 4 instead of 8.
6C1 will fill the final spot (only 6 left here since we filled the first 2 spots with siblings).

8C3 - 4C2*6C1 = 32
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 [#permalink] New post 13 Nov 2007, 11:15
jimjohn wrote:
can someone plz help explain what i did wrong?

there are eight possible choices for the first committee member

for the second committee member, we can't have the first person's sibling on the team so we have six possible choices for the second committee member

for the third committee member we can't have either of the first two chosen members's siblings, so we have four possible choices for the third committee member

so the total num of ways = 8*6*4


Ah ha ! I knew I couldnt have been the only person who thought of it like this !

Scroll up to view my posts, along with jbs'. We had a mini conversation about how to make this approach work, and why it wouldnt work as 8 x 6 x 4
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 [#permalink] New post 13 Nov 2007, 23:41
I understand that there are 4 pairs of siblings, but how can 4C2 = 4? Am I missing something here?
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Re: PS - Combinatorics [#permalink] New post 14 Nov 2007, 06:19
yuefei wrote:
pmenon wrote:
beckee, why 4C2 and 6C1 ? I mean, can you break down how you came up with those #s ?


4C2*6C1 is the total number of combinations with 2 siblings.

There are 3 people in each group.
4C2 is the number of sibling combinations for the first 2 spots in the group. Since we are looking for a pair instead of individual brothers & sisters, we are selecting from 4 instead of 8.
6C1 will fill the final spot (only 6 left here since we filled the first 2 spots with siblings).

8C3 - 4C2*6C1 = 32


Isn't 8c3 - 4c2* 6c1 = 56 - 6*6 = 56-36= 24? How is the OA 32???
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Re: PS - Combinatorics [#permalink] New post 14 Nov 2007, 08:32
Skewed wrote:
yuefei wrote:
pmenon wrote:
beckee, why 4C2 and 6C1 ? I mean, can you break down how you came up with those #s ?


4C2*6C1 is the total number of combinations with 2 siblings.

There are 3 people in each group.
4C2 is the number of sibling combinations for the first 2 spots in the group. Since we are looking for a pair instead of individual brothers & sisters, we are selecting from 4 instead of 8.
6C1 will fill the final spot (only 6 left here since we filled the first 2 spots with siblings).

8C3 - 4C2*6C1 = 32


Isn't 8c3 - 4c2* 6c1 = 56 - 6*6 = 56-36= 24? How is the OA 32???


It should be 8C3 - 4C1*6C1

4C1 because you're picking one pair from 4 pairs.
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 [#permalink] New post 14 Nov 2007, 10:38
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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
ANSWER: C
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 [#permalink] New post 18 Nov 2007, 22:40
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.
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 [#permalink] New post 01 Dec 2007, 09:32
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C.

4C3*2C1*2C1*2C1=4*2^3=32
4C3 - 3 pairs from 4 pairs.
2C1 - changing between brother and sister in each pair.
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Re: PS - Combinatorics [#permalink] New post 26 Aug 2008, 13:23
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


= 8C1*6C1*4C1/3! = 32
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Re: PS - Combinatorics [#permalink] New post 26 Aug 2008, 13:30
4 C 3 * 2 C 1 * 2 C 1 * 2 C 1 = 4*2*2*2 = 32
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Re: PS - Combinatorics (m02q05) [#permalink] New post 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.
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Re: PS - Combinatorics (m02q05) [#permalink] New post 23 Aug 2009, 11:47
Nevermind i got it: The operating phrase here "excluded possibilities!"

in otherwords what cannot be.

The first two people are indeed siblings of whom you ignore to grab someone else from the the remaining six to form a committee that cannot be.

subtracting the the committees that cannot be from the sum of those that can and cannot be leaves you with those that can be.

I hate that it take me so stinking long to figure these out!

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Re: [#permalink] New post 23 Aug 2009, 22:07
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walker wrote:
C.

4C3*2C1*2C1*2C1=4*2^3=32
4C3 - 3 pairs from 4 pairs.
2C1 - changing between brother and sister in each pair.


Right. same as mine.

A(_ _) B(_ _) C(_ _) D(_ _)

from A B C D, choose 3, 4C3
from the chosen 3, pick one from each 2C1*2C1*2C1
in total: 4C3*2C1*2C1*2C1=32
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Re:   [#permalink] 23 Aug 2009, 22:07
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