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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

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.

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.

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

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

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

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)

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? _________________

Cheers! JT........... If u like my post..... payback in Kudos!!

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Re: There are four distinct pairs of brothers and sisters. In [#permalink]

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08 Jul 2013, 17:42

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 _________________

Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

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. _________________

Encourage cooperation! If this post was very useful, kudos are welcome "It is our attitude at the beginning of a difficult task which, more than anything else, will affect It's successful outcome" William James

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

Re: There are four distinct pairs of brothers and sisters. In [#permalink]

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29 Apr 2015, 23:47

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