Simple combination problem

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Resilient
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Simple combination problem

Postby Resilient » Thu May 06, 2010 5:45 am UTC

I was helping someone with some highschool algebra and got myself confused.

So, let there be a test with 10 questions.

You have to answer exactly 7 questions.

Of the first 6 questions, you must answer at least 3 of them.

How many ways are there to answer the questions?

I take this problem to mean that order doesn't matter.

I would use 6 choose 3 to satisfy the second condition. That leaves 7 questions to pick from. So then would I add 7 choose 4? Or would I multiply by 7 choose 4? It seems that I would multiply, but the number is far too large. But, it seems for any 3 picked to satisfy the first condition, there are 7 choose 4 ways to finish it off, which seems like i should be multiplying.

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phlip
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Re: Simple combination problem

Postby phlip » Thu May 06, 2010 5:51 am UTC

You would multiply, except note that it means you're going to count "Answer questions 1 through 7" many times... eg once where 1, 2 and 3 are the ones you choose to satisfy the "at least 3 of 6" constraint, and then 4, 5, 6, and 7 for the remainder... and then again where 1, 2, and 4 are chosen in the first step, and so on. So that's not the way to go.

First, I'll ask: are you sure you got numbers in the question right? Can you give an example of a set of questions to answer where you're answering 7 of the 10 questions, but not answering at least 3 of the first 6?

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Resilient
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Re: Simple combination problem

Postby Resilient » Fri May 07, 2010 2:00 am UTC

No, totally not sure I got the numbers right.

Now that you point that out, I am fairly sure I got the numbers wrong. :)

It was probably something like:

answer exactly 5 questions
and
answer at least 3 of the first 5.

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imatrendytotebag
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Re: Simple combination problem

Postby imatrendytotebag » Fri May 07, 2010 7:09 am UTC

The easiest way I can see to do this, to make absolute sure you don't double-count anything, is to break it up into cases. Using your more recent problem:

Case 1: Answer 3 of the first 5, 2 of the last 5
Case 2: Answer 4 of the first 5, 1 of the last 5
Case 3: Answer 5 of the first 5, 0 of the last 5

Then find the combinations for each case and add em up.

This may also be the best way of doing the problem for higher numbers too.
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