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# There are four distinct pairs of brothers and sisters. In how many way

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SVP
Joined: 21 Jan 2007
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Location: New York City
There are four distinct pairs of brothers and sisters. In how many way  [#permalink]

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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?
Math Expert
Joined: 02 Sep 2009
Posts: 65765
Re: There are four distinct pairs of brothers and sisters. In how many way  [#permalink]

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08 Jul 2013, 21:41
2
5
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?
$$C^2_4*2^2=24$$.

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Hope it helps.
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Re: There are four distinct pairs of brothers and sisters. In how many way  [#permalink]

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11 Mar 2009, 22:28
4
2
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.
##### General Discussion
Intern
Joined: 28 Aug 2007
Posts: 37
Re: There are four distinct pairs of brothers and sisters. In how many way  [#permalink]

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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.

SVP
Joined: 21 Jan 2007
Posts: 1856
Location: New York City
Re: There are four distinct pairs of brothers and sisters. In how many way  [#permalink]

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16 Dec 2007, 21:28
1
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
VP
Joined: 07 Nov 2007
Posts: 1060
Location: New York
Re: There are four distinct pairs of brothers and sisters. In how many way  [#permalink]

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25 Aug 2008, 13:16
4
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?

$$= (8C1*6C1*4C1*2C1)/ 4!$$
$$= 16$$
Intern
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Posts: 3
Re: There are four distinct pairs of brothers and sisters. In how many way  [#permalink]

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28 Aug 2009, 03:25
5
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
Manager
Joined: 27 Oct 2008
Posts: 125
Re: There are four distinct pairs of brothers and sisters. In how many way  [#permalink]

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27 Sep 2009, 20:58
1
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
Intern
Joined: 11 Sep 2009
Posts: 46
Re: There are four distinct pairs of brothers and sisters. In how many way  [#permalink]

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13 Dec 2009, 02:41
3
2
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)
Manager
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Location: United States (VA)
Concentration: Entrepreneurship, Finance
GMAT 1: 680 Q49 V34
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Re: There are four distinct pairs of brothers and sisters. In how many way  [#permalink]

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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
Math Expert
Joined: 02 Sep 2009
Posts: 65765
Re: There are four distinct pairs of brothers and sisters. In how many way  [#permalink]

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14 Aug 2016, 04:42
ashima09 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?

Since the committee shouldn't have siblings in it, then a pair can send only one "representative" to the committee. We have 4 pairs of siblings each can send only one representative: 2*2*2*2 =2^4 = 16.

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Re: There are four distinct pairs of brothers and sisters. In how many way   [#permalink] 14 Aug 2016, 04:42