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

Question

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?

Bunuel · Accepted Answer

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.

csabiga · Answer

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

antihero · Answer

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?

bmwhype2 · 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

x2suresh · Answer

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

alpha_plus_gamma · Answer

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  
+
2 sisters and 2 brothers = 4C2* 2C2 = 6  
+
3 sisters and 1 brother = 4C3* 1C1 = 4  
+
4 sisters and 0 brothers = 4C4 = 1

= 16 total ways.

srivas · Answer

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

Prometoh · Answer

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)

Maxirosario2012 · Answer

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

Bunuel · Answer

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

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

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