# Principle of counting problem

How many committees of 5 can be chosen from 10 people if each committee includes Alice and excludes Bob?

How many committees of 5 can be chosen from 12 people if each committee (i) includes Alice; (ii) excludes Bob; (iii) includes Alice and excludes Bob; (iv) includes at least one of either Alice or Bob

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1) To form a committee of $5$ people which includes Alice, we need to choose just $4$ more members. Since it excludes Bob, we only have $8$ out of the $10$ people to choose from (Alice and Bob have been accounted for). So this can be done in ${8}\choose{4}$ ways.

2)Inclusion into a committee is to be regarded as one less person to choose and exclusion from a committee should be thought as one less person to choose from. The various cases are just a reapplication of these ideas once or more.

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Great mod. I gave you an upvote before seeing this because you had it right. Hope this helps OP. I had tried to comment on mine, but the edits didn't take because of timing. – Ross Millikan Nov 20 '10 at 4:23

@Timothy Wagner: here is my suggestion. Thanks for not doing the hardest one. There is a wide range in how people respond to homework questions, so don't take mine as gospel-make up your own mind. But the differences between us are small. It is in how big a hint to give.

I bet this is homework. Please tag as such.

1)You have 10 people to choose from. Alice is required and Bob is excluded. So there are really only 8 to think about.

2)iii is the same (with 10->12) as the first. i and ii and not so different

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