ParthKohli
  • ParthKohli
Hey.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
ParthKohli
  • ParthKohli
Why is the number of solutions to the following equation:\[x_1 + x_2 + \cdots + x_k=n\]given \(a_i
imqwerty
  • imqwerty
number of solutions =\[x_{i} \]?
ParthKohli
  • ParthKohli
Number of solutions = ordered pairs.

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ParthKohli
  • ParthKohli
@ganeshie8
ParthKohli
  • ParthKohli
yeah
ParthKohli
  • ParthKohli
ah, kinda makes sense to me now.
imqwerty
  • imqwerty
ahhgh that given expression is so bad..
ParthKohli
  • ParthKohli
I understood why. Now as an example if we want to calculate solutions of\[x_1 + x_2 +x_3=20\]subject to \(x_i \ge -3\) then\[(x^{-3} + x^{-2} + \cdots + x^{26})^3\]coefficient of \(x^{20}\)
ParthKohli
  • ParthKohli
\[\left(x^{-3}\cdot \left(\frac{1 - x^{30}}{1 - x}\right)\right)^{3}\]
ParthKohli
  • ParthKohli
^ how do we calculate the coefficient of \(x^{20}\) in that?
imqwerty
  • imqwerty
i didn't get that 2nd step? (x^-3 +x^-2..x^26)^3
ParthKohli
  • ParthKohli
ah, the lower restriction is given to us: \(x_1, x_2, x_3 \ge -3\) since their sum is \(20\), the max. value any of them can take is \(26\) (if and only if the other two are -3) thus \(-3 \le x_i \le 26\)
ParthKohli
  • ParthKohli
now refer to the above identity
imqwerty
  • imqwerty
:o ok now i get the ques :)
ParthKohli
  • ParthKohli
but how do we calculate the coefficient? :(
ParthKohli
  • ParthKohli
ah, since \(1 -x^{30}\) cannot generate \(x^{20}\) we can remove that and find the coefficient of \(x^{20}\) in \(x^{-9} \cdot \dfrac{1}{(1-x)^3}\)
ParthKohli
  • ParthKohli
I'm not sure about the above step... can you confirm it?
imqwerty
  • imqwerty
wait what was that equation where did it go?http://prntscr.com/8nkh3q
ParthKohli
  • ParthKohli
maybe there are displaying problems... refresh?
ganeshie8
  • ganeshie8
generating functions

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