venmic wrote:If there are four distinct pairs of brothers and sisters, then in how many ways can a committee of 3 be formed and NOT have siblings in it?
8
24
32
56
192
What is the logic in this question
56
Hi --I think the answer you provided is Wrong. Read the post...
This question is similar to the one of the MGMAT question discussed on the Manhattan forum.. That question was about choosing 3 cars each of different color.
Here is the link --
https://www.manhattangmat.com/forums/the ... 12659.html
Now, Lets look at this question... First make 4 pairs
B1S1 B2S2 B3S3 B4S4
Select the first person from the group of 8 people ( 4 pair = 4*2)
This can be done in
8 WAYS, say B1 was selected.
Now, for the next selection S1 cannot participate ..(Reason -- constraint given in the question says that siblings cannot be together)
Therefore, we have only 6 people to choose from.
2nd person can be chosen in
6 WAYS.
Similarly, 3rd person can be chosen in
4 ways..
Total = 8*6*4 --
is this the answer? [WAIT]
We have to see that the ORDER of selection does not matter. So we divide the product by 3! [3! because we chose 3 people]
The correct answer to this question will be -- [spoiler](8*6*4)/3! = 32[/spoiler]