Combinatorics q
Have I done b right and how do I do c?
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Original post by cooldudeman
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.
(edited 9 years ago)
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?
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?
Spoiler
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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Original post by cooldudeman
WHts the difference between nonnegative and positive solutions?
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.
(edited 9 years ago)
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.
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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(edited 9 years ago)
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.
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.
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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Original post by cooldudeman
Yes.
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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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.
Original post by ghostwalker
OK
Huh! What's part b got to do with it?
Think about what your A and B are here.
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?
Original post by cooldudeman
So its just 16C15 + 9C8? That simple?
Almost. You need to subtract |AnB| as per your formula in part a.
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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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.
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.
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
Original post by MSI_10
I got 18 Choose 16 as well by following the notes as attached
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.
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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