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.

## Simple combination problem

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- phlip
- Restorer of Worlds
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### Re: Simple combination problem

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?

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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void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

### Re: Simple combination problem

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.

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.

- imatrendytotebag
**Posts:**152**Joined:**Thu Nov 29, 2007 1:16 am UTC

### Re: Simple combination problem

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.

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.

Hey baby, I'm proving love at nth sight by induction and you're my base case.

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