cooldudeman
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Have I done b right and how do I do c?

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ghostwalker
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(Original post by cooldudeman)
Have I done b right
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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cooldudeman
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(Original post by 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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On my notes it has this example which they subbed in y. I dont understand why. WHts the difference between nonnegative and positive solutions?

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ghostwalker
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(Original post by cooldudeman)
WHts the difference between nonnegative and positive solutions?
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.
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cooldudeman
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(Original post by 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.
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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ghostwalker
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(Original post by cooldudeman)
Is the answer 21choose3?
No - read your corollary carefully.

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

Would it be 21choose3 - 16choose2 - 9choose2?
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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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cooldudeman
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(Original post by 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.
Ohh 21C19? Gonna try next part now.

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ghostwalker
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(Original post by cooldudeman)
Ohh 21C19? Gonna try next part now.

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Yes.
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cooldudeman
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(Original post by ghostwalker)
Yes.
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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ghostwalker
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(Original post by 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.
OK

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

Think about what your A and B are here.
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cooldudeman
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(Original post by ghostwalker)
OK



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

Think about what your A and B are here.
So its just 16C15 + 9C8? That simple?
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ghostwalker
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(Original post by cooldudeman)
So its just 16C15 + 9C8? That simple?
Almost. You need to subtract |AnB| as per your formula in part a.
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cooldudeman
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(Original post by ghostwalker)
Almost. You need to subtract |AnB| as per your formula in part a.
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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ghostwalker
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(Original post by 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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Yep, and you're done.
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MSI_10
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(Original post by 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.
I got 18 Choose 16 as well by following the notes as attached
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cooldudeman
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(Original post by MSI_10)
I got 18 Choose 16 as well by following the notes as attached
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Yeah I was confused before with this. Its the terms nonnegative(all natural numbers including zero) and positive(0 is not a positive nunber). They are different. The example on the lecture notes says positive so the you have to use the substitution y1=x1-1 etc. x1 can be 1 at the minimum so y1 can be zero in this case.

The question in the sample test says nonnegative so you just use the corollary without anything else.
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ghostwalker
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(Original post by MSI_10)
I got 18 Choose 16 as well by following the notes as attached
As cooldudeman said in previous post.
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