Anonymous

# Combination related

How many committees of 5 can be chosen from Western's 13-member Leadership Team if Mr.Brown and Mr.White are not allowed to serve on the same committee?

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Lv 6

So you have three scenarios:

A: both Mr. Brown and Mr. White are NOT on the committee:

Then there are 11 people for 5 spots, that's 11 choose 5 or (11! / (5! * 6!)) = 462

B: Mr. Brown is on the committee but not Mr. White:

There are 4 spots left (Mr. Brown took one spot), but 11 people left to choose from (excludes Mr. White). This is 11 people for 4 spots, so 11C4 = 330.

C: Mr. Brown is not on the committee but Mr. White is: also 330, same reasoning as B.

Add them all together = 1122.

Alternatively, you can think of it like this. The total number of ways = all possible ways to make a 5 member committee from 13 people - the number of ways that both Mr. Brown and Mr. White are on the committee.

This equals 13C5 - 11C3 = 1287 - 162 = 1122.

Source(s): I'm a graduate student in Genetics... so we had to learn probability/permutation/combination.