This week, we have a guest post written by Alexander B. on generating functions, which are useful tools for solving advanced counting problems.

How would you use generating functions to solve the following?

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How many non-negative integer solutions are there to \( a + b + 2c = n \) for any positive integer \(n\)?

Share a question which can be approached using Generating Functions.

If you are interested in contributing a post, please contact me at Calvin@Brilliant.org.

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## Comments

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TopNewestAfter you finish number 1, try this:

How many non-negative integer solutions are there to \(a+b+2c=n\) for any positive integer \(n\)

for distinct \(a, b, c\)?Log in to reply

If \(d_n\) is the number of solutions with \(a=b\), then \[ d_n \; = \; \left\{ \begin{array}{lcl} \tfrac12n+1 & \quad & n \mbox{ even} \\ 0 & \quad & n \mbox{ odd} \end{array}\right. \] If \(e_n\) is the number of solutions with \(a=c\) or \(b=c\) then \[ e_n \; = \; 1 + \big\lfloor \tfrac13n\big\rfloor \] If \(f_n\) is the number of solutions with \(a=b=c\) then \[ f_n \; = \; \left\{ \begin{array}{lcl} 1 & \quad & 4\,|\,n \\ 0 & \quad & 4 \not| n \end{array} \right. \] and the number of solutions with \(a,b,c\) distinct is \[ c_n - d_n - 2e_n + 2f_n \] where \(c_n = \big\lfloor \tfrac14(n+2)^2\big\rfloor\) is as in my other post.

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The number of solutions, \(c_n\), is the coefficient of \(x^n\) in the expansion of \[ \frac{1}{(1-x)^2(1-x^2)} \; = \; \frac{1}{(1-x)^3(1+x)} \; = \; \left(\sum_{j=0}^\infty {j+2 \choose 2}x^j\right)\left(\sum_{k=0}^\infty (-1)^kx^k\right) \] and so \[ \begin{array}{rcl} c_n & = & \sum_{m=0}^n (-1)^{n-m}{m+2 \choose 2} \; = \; \tfrac12(-1)^n\sum_{m=0}^n(-1)^m(m+1)(m+2) \\ & = & \tfrac14(-1)^n\sum_{m=0}^n (-1)^m\big[(m+1)^2+(m+2)^2-1\big] \\ & = & \tfrac14(-1)^n\left[\sum_{m=0}^n (-1)^m(m+1)^2 - \sum_{m=1}^{n+1}(-1)^m(m+1)^2 - \sum_{m=0}^n(-1)^m\right] \\ & = &\tfrac14(-1)^n\left[1 + (-1)^n(n+2)^2 - \sum_{m=0}^n(-1)^m\right] \; = \; \tfrac14\big[(n+2)^2 - \varepsilon_n\big] \end{array} \] where \[ \varepsilon_n \; = \; \left\{ \begin{array}{lcl} 0 & \qquad & n \mbox{ even} \\ 1 & \qquad & n \mbox{ odd} \end{array} \right. \] and hence \[ c_n \; = \; \left \lfloor \tfrac14(n+2)^2 \right\rfloor \]

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Awesome..Till now, I was completely unknown to this concept,(I had seen it in problems but never understood )but now I am very interested in it...I am very happy now..thnx Alexander and Calvin.

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Can you use generating functions when you are picking 2 or more

distinctelements from one set?Log in to reply

Consider \( f(x)^2 - f(x^2) \).

What would it look like for picking 3 distinct objects?

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Ah, brilliant. Though as you start picking 3, 4, 5, etc. elements this doesn't become very practical anymore.

To your edit-- It would be \(f(x)^3 - f(x^3) - \) some other things to get rid of the products in which only a pair are the same. I'll think about it.

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I don't think your example works.

Hint:Just because you learnt an advanced technique, doesn't mean that you should forget your basics.Log in to reply

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Principle of Inclusion and Exclusion.

To elaborate on my hint, use theThis tells you why the formula looks the way it does (and also why it gets ugly quickly). Interestingly, it only involves terms of the form \( f(x^n) \), so if you have a simple description of \( f(x) \), then you might be able to get a simple description for the generating function of distinct elements.

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If \(f(x) = x^4 + x^5 + x^6 + x^7\) then it would be \((f(x))^3 - x^4 f(x^2) - x^5 f(x^2) - x^6f(x^2) - x^7f(x^2)\)?

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Your formula looks like \( f(x)^3 - f(x) f(x^2) \).

If it works, why does it work?

If it doesn't work, why doesn't it work?

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