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?

I thought so looks a lot like Challenge 2 question 5

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

I understand why is 384 upside....but why 4! downside? please help me

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.

suresh, could you please elaborate your calculation?

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

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