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There are four distinct pairs of brothers and sisters. In

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There are four distinct pairs of brothers and sisters. In [#permalink] New post 02 Dec 2007, 13:57
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There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?
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Re: combinatorics - siblings [#permalink] New post 02 Dec 2007, 23:22
bmwhype2 wrote:
There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?


I get 4. 8!/4!4! = 70

Then we have
BS12, BS13, BS14, BS15, BS16 1's done

BS23, BS24, BS25, BS26 2's

BS34, BS35, BS36

BS45, BS46

BS56

total 15 ways. So 15*4=60 b/c total of 4 sibling pairs.


Now we have to count BSBS amounts. 4!/2!2! = 6.

so 70-66=4

I have no idea if this is right though. Is this a made up problem or an official problem?
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 [#permalink] New post 04 Dec 2007, 01:37
8C4 - (4x6C2 - 4C2)

Explanation

8C4 includes all the combinations = 70
4x6C2 includes all the combinations which is turn includes all the two pairs twice i.e. A1A2B1B2 and B1B2A1A2
so only need to subtract once the 4C2

In this scenario brute force seems easier. I don't think i would have figured out to subtract only once the 4C2 it in the GMAT under pressure
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Re: combinatorics - siblings [#permalink] New post 04 Dec 2007, 06:26
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bmwhype2 wrote:
There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?


8*6*4*2 / 4! = 384/24 = 16

Just brute force it, on first place you can put 8, on second you can put 6 (excluding 1 sibling) on third you can put 4 (exclude) 2 siblings... then divide by the number of permutations as position doesnt matter.

Whats the answer?
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Re: combinatorics - siblings [#permalink] New post 04 Dec 2007, 07:53
antihero wrote:
bmwhype2 wrote:
There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?


8*6*4*2 / 4! = 384/24 = 16

Just brute force it, on first place you can put 8, on second you can put 6 (excluding 1 sibling) on third you can put 4 (exclude) 2 siblings... then divide by the number of permutations as position doesnt matter.

Whats the answer?


yeap, the answer is 16. you can find the exact same question in challenges (http://www.gmatclub.com/tests/m/02/#q5) but with the committee of 3
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 [#permalink] New post 16 Dec 2007, 21:28
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i'm not even sure of the answer. maybe walker can help out.

4 pairs = 4*2 = 8 people total

8C4 = 8!/4!4! = 70 total outcomes

Total – unfavorable = favorable

Unfavorable outcomes
Assuming one pair of twins in the committee, we have two spaces left. Since we plugged a pair of twins in the committee, we have 8-2= 6 people to fill 2 spaces.

6C2 = 6!/2!4! = 15 ways to fill the two remaining slots

We only filled the slots with one pair, and we have to account for arrangements of the pairs. Now, we have 4 pairs. 4*15= 60 total arrangements

When we place members into the remaining slots, there may be an additional set of twins. There are 2 remaining slots to which we can fit a pair of twins. If it were one remaining slot, we cannot fit a pair of twins, so we wouldn’t have to account for duplicates.

Now we account for the number of duplicates.
# of duplicates = Total arrangements - # of unique combinations

To find the # of duplicates of twins, we need treat a pair of twins as one unit.
This means the 4 slots are really 2 slots.

Total ways of arranging four pairs of twins in two slots
4P2 = 4*3 = 12 total ways

Total # of unique combinations
Choosing two pairs out of 4 pairs
= 4C2
= 6
Therefore, # of duplicates = 12 - 6 = 6 duplicates

70 – 60 + 6 = 16
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Re: combinatorics - siblings [#permalink] New post 25 Aug 2008, 13:16
bmwhype2 wrote:
There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?


= (8C1*6C1*4C1*2C1)/ 4!
= 16
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Re: combinatorics - siblings [#permalink] New post 11 Mar 2009, 22:28
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another method

The commitee of 4 , NOT having siblings can be formed in following ways:

0 Sisters and 4 brothers = 4C4 =1
+
1 sister and 3 brothers = 4C1*3C3 = 4 [ 3C3 because selected sister's brother can not be among 3 bros]
+
2 sisters and 2 brothers = 4C2* 2C2 = 6 [ again 2C2 because brothers of 2 selected sisters can not be on commitee]
+
3 sisters and 1 brother = 4C3* 1C1 = 4 [ only one brother whose sister is not on commitee can be selected]
+
4 sisters and 0 brothers = 4C4 = 1

= 16 total ways.
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Re: combinatorics - siblings [#permalink] New post 28 Aug 2009, 03:25
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The four distinct pairs of brothers and sisters:

Aa Bb Cc Dd

We have to choose one from each set. We can do it:
2 * 2 * 2 * 2 = 16 ways
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Re: combinatorics - siblings [#permalink] New post 28 Aug 2009, 04:14
I have another result...could someone tell me where my logic is false?

Total possible combinations is 70
there are only 4 combinations possible for siblings B1S1, B2S2, B3S3 and B4S4

so for me there are 70-4 = 66

Can someone comment, correct?
Thx
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Re: combinatorics - siblings [#permalink] New post 28 Aug 2009, 04:35
I approached the question in the following way

Let's say that we have four pairs = {Aa,Bb,Cc,Dd} capital letters represent the boys and small letters the girls, no offense.

A committee will be formed just by one member of each pair so the committee will be

A or a for the first person, B or b for the second and so on
since you have two choices for each position 2*2*2*2 = 16 ways
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Re: combinatorics - siblings [#permalink] New post 04 Sep 2009, 22:47
To make a committee consist of 4 member out of 4 pair of siblings .lets say A1A2 B1B2 D1D2 E1E2
we have to select one out of each sibling pair
no of selection for each pair =2c1=2
we need to select this way from each pair=2C1 (to select from A1A2) * 2C1 (to select from B1B2)*2C1 (D1D2) *2C1(E1E2)
=2C1^4=2^4=16 :-D
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Re: combinatorics - siblings [#permalink] New post 27 Sep 2009, 20:58
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There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?

Soln:
2C1 * 2C1 * 2C1 * 2C1 = 16 ways
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Re: combinatorics - siblings [#permalink] New post 13 Dec 2009, 02:41
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8 * 6 * 4 *2 / 4! = 16

Note that 8 * 6 * 4 *2 create a duplicate such as ABCD and BACD. Thus, we need to cancel out the duplicates by dividing 4! (there are 4! ways to shuffle the committee)
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Re: combinatorics - siblings [#permalink] New post 16 Feb 2010, 09:11
x2suresh wrote:
bmwhype2 wrote:
There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?


= (8C1*6C1*4C1*2C1)/ 4!
= 16


I could understand.. 8C1*6C1*4C1*2C1.. But I am not able to place.. why we divide by 4!??? Can anyone help?
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Re: There are four distinct pairs of brothers and sisters. In [#permalink] New post 08 Jul 2013, 16:42
What would be the ansmer if instead of a committee of 4 we would need a committee of 3?
64 possible committees?

2^4 * C^4_3 = 16*4 = 64

And a committee of 2? 96?
2^4 * C^4_2 = 16*6 = 96
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Re: There are four distinct pairs of brothers and sisters. In [#permalink] New post 08 Jul 2013, 21:41
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Maxirosario2012 wrote:
What would be the ansmer if instead of a committee of 4 we would need a committee of 3?
64 possible committees?

2^4 * C^4_3 = 16*4 = 64

And a committee of 2? 96?
2^4 * C^4_2 = 16*6 = 96


There are four distinct pairs of brothers and sisters.

A. In how many ways can a committee of 4 be formed and NOT have siblings in it?
2^4

B. In how many ways can a committee of 3 be formed and NOT have siblings in it?
C^3_4*2^3=32.
Check here: if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html

C. In how many ways can a committee of 2 be formed and NOT have siblings in it?
C^2_4*2^2=24.

Similar questions to practice:
in-a-room-filled-with-7-people-4-people-have-exactly-87550-20.html
a-certain-junior-class-has-1-000-students-and-a-certain-58914.html
if-there-are-four-distinct-pairs-of-brothers-and-sisters-99992.html
a-dog-breeder-currently-has-9-breeding-dogs-6-of-the-dogs-131992.html
three-pairs-of-siblings-each-pair-consisting-of-one-girl-136837.html

Hope it helps.
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Re: [#permalink] New post 16 Jul 2013, 15:32
bmwhype2 wrote:
i'm not even sure of the answer. maybe walker can help out.

4 pairs = 4*2 = 8 people total

8C4 = 8!/4!4! = 70 total outcomes

Total – unfavorable = favorable

Unfavorable outcomes
Assuming one pair of twins in the committee, we have two spaces left. Since we plugged a pair of twins in the committee, we have 8-2= 6 people to fill 2 spaces.

6C2 = 6!/2!4! = 15 ways to fill the two remaining slots

We only filled the slots with one pair, and we have to account for arrangements of the pairs. Now, we have 4 pairs. 4*15= 60 total arrangements

When we place members into the remaining slots, there may be an additional set of twins. There are 2 remaining slots to which we can fit a pair of twins. If it were one remaining slot, we cannot fit a pair of twins, so we wouldn’t have to account for duplicates.

Now we account for the number of duplicates.
# of duplicates = Total arrangements - # of unique combinations

To find the # of duplicates of twins, we need treat a pair of twins as one unit.
This means the 4 slots are really 2 slots.

Total ways of arranging four pairs of twins in two slots
4P2 = 4*3 = 12 total ways

Total # of unique combinations
Choosing two pairs out of 4 pairs
= 4C2
= 6
Therefore, # of duplicates = 12 - 6 = 6 duplicates

70 – 60 + 6 = 16


Very interesting explanation. But I could not understand why we have to subtract the duplicates and how are you calculating them.
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Re: combinatorics - siblings [#permalink] New post 16 Sep 2013, 09:10
:-D very nice explanation


The commitee of 4 , NOT having siblings can be formed in following ways:

0 Sisters and 4 brothers = 4C4 =1
+
1 sister and 3 brothers = 4C1*3C3 = 4 [ 3C3 because selected sister's brother can not be among 3 bros]
+
2 sisters and 2 brothers = 4C2* 2C2 = 6 [ again 2C2 because brothers of 2 selected sisters can not be on commitee]
+
3 sisters and 1 brother = 4C3* 1C1 = 4 [ only one brother whose sister is not on commitee can be selected]
+
4 sisters and 0 brothers = 4C4 = 1

= 16 total ways.[/quote]
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Re: combinatorics - siblings [#permalink] New post 24 Sep 2013, 05:22
x2suresh wrote:
bmwhype2 wrote:
There are four distinct pairs of brothers and sisters. In how many ways can a committee of 4 be formed and NOT have siblings in it?


= (8C1*6C1*4C1*2C1)/ 4!
= 16


Hi Suresh,
Could you please explain your reasoning?

Thanks
Re: combinatorics - siblings   [#permalink] 24 Sep 2013, 05:22
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