# Combinatorics q

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Have I done b right

**cooldudeman**)Have I done b right

You can actually do this easily the "long" way. If x1=0, there are 20 possible solutions, if x1=1, there are 19, etc. and sum.

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(Original post by

No, check your notes. If you put yi = xi-1, then when x1 is 0, as it can be, y1 will be -1; does that seem reasonable?

You can actually do this easily the "long" way. If x1=0, there are 20 possible solutions, if x1=1, there are 19, etc. and sum.

If x1 and x2 are defined, what must x3 be to satisfy your constraint?

Hence the number of solutions must be?

**ghostwalker**)No, check your notes. If you put yi = xi-1, then when x1 is 0, as it can be, y1 will be -1; does that seem reasonable?

You can actually do this easily the "long" way. If x1=0, there are 20 possible solutions, if x1=1, there are 19, etc. and sum.

If x1 and x2 are defined, what must x3 be to satisfy your constraint?

Hence the number of solutions must be?

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#4

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WHts the difference between nonnegative and positive solutions?

**cooldudeman**)WHts the difference between nonnegative and positive solutions?

Positive integers: 1,2,3,....

Non-negative solutions use the first set, and positive solutions use the second set.

You want "Corollary 2.5" at the top, as it's stated.

PS: Last post for today.

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(Original post by

Non-negative integers: 0,1,2,3,4,....

Positive integers: 1,2,3,....

Non-negative solutions use the first set, and positive solutions use the second set.

You want "Corollary 2.5" at the top, as it's stated.

PS: Last post for today.

**ghostwalker**)Non-negative integers: 0,1,2,3,4,....

Positive integers: 1,2,3,....

Non-negative solutions use the first set, and positive solutions use the second set.

You want "Corollary 2.5" at the top, as it's stated.

PS: Last post for today.

If it is, could you tell me how to do pay c please.

Would it be 21choose3 - 16choose2 - 9choose2?

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#6

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Is the answer 21choose3?

**cooldudeman**)Is the answer 21choose3?

If it is, could you tell me how to do pay c please.

Would it be 21choose3 - 16choose2 - 9choose2?

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Would it be 21choose3 - 16choose2 - 9choose2?

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I've deleted my original response to part c, as I didn't read the question properly.

Part c), utilises the inclusion/exclusion property.

How many solutions are there for x1=4? Write them all out if necessary.

Similarly x2=11.

And "x1=4 and x2=11".

And include/exclude, so to speak.

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(Original post by

No - read your corollary carefully.

No.

I've deleted my original response to part c, as I didn't read the question properly.

Part c), utilises the inclusion/exclusion property.

How many solutions are there for x1=4? Write them all out if necessary.

Similarly x2=11.

And "x1=4 and x2=11".

And include/exclude, so to speak.

**ghostwalker**)No - read your corollary carefully.

No.

I've deleted my original response to part c, as I didn't read the question properly.

Part c), utilises the inclusion/exclusion property.

How many solutions are there for x1=4? Write them all out if necessary.

Similarly x2=11.

And "x1=4 and x2=11".

And include/exclude, so to speak.

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(Original post by

Yes.

**ghostwalker**)Yes.

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#10

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If x1 is 4 the we have x2+x3=15 which there are 16C15 and if x2=11 then x1+x2=8 so 9C8.

**cooldudeman**)If x1 is 4 the we have x2+x3=15 which there are 16C15 and if x2=11 then x1+x2=8 so 9C8.

For part b we have the |AnB| which is more than the sum of them two. Do you use the formula for three sets or two? With two sets, it doesnt work.

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Think about what your A and B are here.

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(Original post by

OK

Huh! What's part b got to do with it?

Think about what your A and B are here.

**ghostwalker**)OK

Huh! What's part b got to do with it?

Think about what your A and B are here.

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#12

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So its just 16C15 + 9C8? That simple?

**cooldudeman**)So its just 16C15 + 9C8? That simple?

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(Original post by

Almost. You need to subtract |AnB| as per your formula in part a.

**ghostwalker**)Almost. You need to subtract |AnB| as per your formula in part a.

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#14

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So if x1 is 4 and x2=11 then there is only one solution. So it would be them two plus then minus one?

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**cooldudeman**)So if x1 is 4 and x2=11 then there is only one solution. So it would be them two plus then minus one?

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#15

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No, check your notes. If you put yi = xi-1, then when x1 is 0, as it can be, y1 will be -1; does that seem reasonable?

You can actually do this easily the "long" way. If x1=0, there are 20 possible solutions, if x1=1, there are 19, etc. and sum.

**ghostwalker**)No, check your notes. If you put yi = xi-1, then when x1 is 0, as it can be, y1 will be -1; does that seem reasonable?

You can actually do this easily the "long" way. If x1=0, there are 20 possible solutions, if x1=1, there are 19, etc. and sum.

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The question in the sample test says nonnegative so you just use the corollary without anything else.

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#17

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I got 18 Choose 16 as well by following the notes as attached

**MSI_10**)I got 18 Choose 16 as well by following the notes as attached

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