Find the number of solutions to the equation :
x1+x2+x3+x4+x5 = 21
where xi is a non negative integer
Provided that
1) 0<=x1<=10
2) x1 is from 0 to 3 , x2 is from 1 to 4, x3 >=15

- DLS

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- DLS

I know that the standard formula for calculating the number of non negative integral solutions is given by \(\Large C(n+r-1,r-1)\). And I know if there are constraints like \(\Large x_i > \alpha\) then we replace \(\Large x_i\) with \(\Large x_i - \alpha\) .(Hopefully I have these facts right). But I am not sure how to deal with these 2 conditions.

- DLS

@ganeshie8

- ganeshie8

Hey you may use generating functions

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## More answers

- DLS

Can you give a good place to learn that ? :/ my textbook has described it in a complicated way :|

- ganeshie8

I have a very good tutorial on generating functions, one sec

- ganeshie8

Here it is
http://db.math.ust.hk/notes_download/elementary/algebra/ae_A11.pdf
@DLS

- ganeshie8

1) 0<=x1<=10
you may proceed like this :
since \(x_1\) can take values from \(0\) to \(10\),
it can be 0 or 1 or 2 or .... or 10
the corresponding factor for \(x_1\) in the generating function is
\((x^0+x^1+x^2+\cdots + x^{10})\)

- ganeshie8

There are no restrictions on other four variables, so the corresponding factor for each of them is
\((x^0+x^1+x^2+\cdots)\)

- ganeshie8

The overall generating function is then given by
\[G(x)= (x^0+x^1+x^2+\cdots + x^{10}) (x^0+x^1+x^2+\cdots )^4\]

- ganeshie8

The answer for part 1 is the coefficient of \(x^{21}\) of \(G(x)\)

- ganeshie8

Can you think of a way to find the coefficient of \(x^{21}\) of \(G(x)\) ?

- DLS

oh this thing is called generating functions? :D didn't know that.
we can use multinomial theorem maybe?

- ganeshie8

we don't really need to get all that sophisticated

- ganeshie8

binomial theorem will do :)

- DLS

how are you planning to apply binomial here :o

- ganeshie8

\[G(x)= (x^0+x^1+x^2+\cdots + x^{10}) (x^0+x^1+x^2+\cdots )^4\]
How are you planning to avoid binomial theorem here ?

- ganeshie8

\[G(x)= (x^0+x^1+x^2+\cdots + x^{10}) (x^0+x^1+x^2+\cdots )^4\\~\\
=(\dfrac{1-x^{11}}{1-x}) (\dfrac{1}{1-x})^4
\]

- ganeshie8

does that look okay to you?

- ganeshie8

I have just used the partial sum and infinite sum formulas of geometric series

- DLS

what did you do in the first term :o

- ganeshie8

\[x^0+x^1+x^2+\cdots + x^{10}\]
Notice that this is a geometric series with first term = \(1\) and common ratio = \(x\)

- ganeshie8

remember the partial sum formula of geometric series ?

- DLS

oh right! sorry,my bad, yeah got that step!

- ganeshie8

good, try simplifying that expression a bit

- DLS

i can proceed now :)

- ganeshie8

familiar with extended binomial theorem for negative exponents ?

- ganeshie8

\[G(x)= (x^0+x^1+x^2+\cdots + x^{10}) (x^0+x^1+x^2+\cdots )^4\\~\\
=(\dfrac{1-x^{11}}{1-x}) (\dfrac{1}{1-x})^4\\~\\
= (1-x^{11})(1-x)^{-5}
\]

- ganeshie8

see if that looks okay

- DLS

yep :)

- DLS

we can open the bracket now i guess

- DLS

\[=>(1-x^{11}) - x^{11}(1-x)^5\]

- DLS

we could apply binomial theorem on the 2nd term :o

- DLS

yeah that's fine too :)

- ganeshie8

\[=>(1-x^{11}) - x^{11}(1-x)^5\]
how did you get that ?

- DLS

sorry the power is -5

- ganeshie8

\[G(x)= (x^0+x^1+x^2+\cdots + x^{10}) (x^0+x^1+x^2+\cdots )^4\\~\\
=(\dfrac{1-x^{11}}{1-x}) (\dfrac{1}{1-x})^4\\~\\
= (1-x^{11})(1-x)^{-5}\\~\\
=(1-x^{11} )(\sum\limits_{k=0}^{\infty}\dbinom{-5}{k} (-x)^k)
\]

- DLS

I meant this..but it isn't required.
\[\Large (1-x^{-5}) - x^{11}(1-x^{-5})\]

- DLS

I thought of writing it like that because the coefficient of x^17 would only come from the 2nd term.

- ganeshie8

do you mean
I meant this..but it isn't required.
\[\Large (1-x)^{-5} - x^{11}(1-x)^{-5}\]
?

- DLS

yeah :P equation editor is trolling me today :/

- ganeshie8

that looks god
as you can see the first term \((1-x)^{-5}\) contains an \(x^{21}\) term
and the second term \(x^{11}(1-x)^{-5}\) also contains an \(x^{21}\) term
we need to find them both

- ganeshie8

whats the coefficient of \(x^{21}\) in the expansion of \((1-x)^{-5}\) ?

- DLS

\[\large C(-5,r)1^r (-x)^{-n-r} => C(-5+r-1,r)(-x)^{-5-r}\]

- DLS

-5-r = 21
r = -26?

- DLS

This is the 2nd or 3rd time I happen to use binomial theorem for negative index :|

- ganeshie8

you're making it way more complicated than it actually is

- DLS

lol sorry

- ganeshie8

whats the coefficient of \(x^{21}\) in the expansion of \((1-x)^{-5}\) ?
the term is : \(C(-5, 21) (-x)^{21}\)

- ganeshie8

therefore the coefficient of \(x^{21}\) is simply \(-C(-5, 21)\)

- ganeshie8

How we evaluate that is another story

- DLS

ohh..all this is because I haven't got the grip of this topic anymore :/ I had done it in XIIth or so, due to the gap in first year I have totally forgotten it :/ sorry..but I happen to remember it now :)

- ganeshie8

its all good :) see if you can find the "required" coefficient in second term

- ganeshie8

\[\Large (1-x)^{-5} - x^{11}\color{red}{(1-x)^{-5}}\]

- ganeshie8

do we need the coefficient of \(x^{10}\) in that red term ?

- DLS

yep..we need all such indices which can add up with 11 to produce 21

- ganeshie8

there is only one number that produces 21 when added to 11

- ganeshie8

and that number is 10

- ganeshie8

once you see how all this works, it is a very simple problem actually...

- DLS

yeah hopefully this would help in clearing alot of questions :|
anyway, how do we get to know which all powers will (1-x)^-5 have ?

- ganeshie8

\((1-x)^{-5} = \sum\limits_{k-0}^{\infty}C(-5, k)(-x)^k\)

- ganeshie8

as you can see it will have all the powers of x

- ganeshie8

we're just looking for the coefficient of \(x^{21}\)

- DLS

I don't know why I'm getting so much confused lol, I get it :D

- ganeshie8

its okay, read the first 3 pages of the pdf i gave you earlier

- DLS

ALRIGHT! so we have the coefficient from the 2nd term now.

- DLS

and the first term as well

- DLS

just a last step to evaluate that term

- ganeshie8

what is the coefficient from second term ?

- DLS

C(-5,10)

- DLS

+C(-5,21) from the first term

- DLS

= total answer

- ganeshie8

Excellent! careful about the signs though

- DLS

oh yes. there is a (-1)^n hanging around there

- DLS

only the first term should have it.

- ganeshie8

\[\Large (1-x)^{-5} - x^{11}\color{red}{(1-x)^{-5}}\]
\(x^{21}\) coefficient from first term is \(-C(-5,21)\)
\(x^{21}\) coefficient from second term is \(C(-5,10)\)
subtract them to get
\[-C(-5,21)-C(-5,10)\]

- ganeshie8

that is our final answer

- DLS

yep!!

- ganeshie8

evaluate that

- ganeshie8

treat that same as a regular binomial coefficient

- ganeshie8

\[C(n,k) = \dfrac{n(n-1)(n-2)\cdots (n-k+1)}{k!}\]

- ganeshie8

\[C(-5,21) = \dfrac{-5(-5-1)(-5-2)\cdots (-5-21+1)}{21!}\]

- ganeshie8

whatever you get

- ganeshie8

it should simplify nicely though

- ganeshie8

lets try

- ganeshie8

\[C(-5,21) = \dfrac{-5(-5-1)(-5-2)\cdots (-5-21+1)}{21!}\\~\\
=\dfrac{(-1)^{21}5(5+1)(5+2)\cdots (5+21-1)}{21!}
\]

- ganeshie8

Oh my! that is same as
\[C(-5,21) = \dfrac{-5(-5-1)(-5-2)\cdots (-5-21+1)}{21!}\\~\\
=\dfrac{(-1)^{21}5(5+1)(5+2)\cdots (5+21-1)}{21!}\\~\\
=(-1)^{21}C(5+21-1,~21)
\]

- ganeshie8

that is just a regular binomial coefficient which you can evaluate using ur calculator

- ganeshie8

That is so awesome! In general do we have below formula to convert negative binomial coefficients to positive :
\[C(-n,~r) = (-1)^r C(n+r-1~r)\]

- ganeshie8

?

- DLS

ohh..nice :D so did you generalize a formula for negative coefficient btw ?
\[\Large (-1)^r C(r-n+1)\]

- ganeshie8

Now I see... you have tried to use that formula earlier right ?

- DLS

haha yes :p

- ganeshie8

Oh my! that is same as
\[C(-5,21) = \dfrac{-5(-5-1)(-5-2)\cdots (-5-21+1)}{21!}\\~\\
=\dfrac{(-1)^{21}5(5+1)(5+2)\cdots (5+21-1)}{21!}\\~\\
=(-1)^{21}C(5+21-1,~21)\\~\\
=-C(25,~21)
\]

- ganeshie8

does that look good ?

- DLS

yeah alot better :D

- ganeshie8

wolfram says the final answer is 11649
http://www.wolframalpha.com/input/?i=-%28-5+choose+21%29-%28-5+choose+10%29

- ganeshie8

il let you finish it off :)

- DLS

yeah I can do the rest ! :D

- DLS

I can probably do the 2nd part of the question too.

- ganeshie8

can you show me the generating function for part2 quick

- ganeshie8

im going for dinner... and i may not login again tonight

- ganeshie8

just want to see what you get for part2

- DLS

2) x1 is from 0 to 3 , x2 is from 1 to 4, x3 >=15
\[\large (x^0+x^1+x^2+x^3)(x^1+x^2+x^3+x^4)(x^{15}+x^{16}...)\]

- DLS

and rest two are infinite GP ..can be whatever

- DLS

i.e 0 to infinity..so 2 more brackets multiplied by that

- ganeshie8

Excellent!
Looks you have nailed these... gtg.. have fun :)

- DLS

BIG THANKS TO YOU!! thanks for your precious time :")

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