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

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?

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

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

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.

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

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)

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

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.

Answer: 16.

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