Generating function

Question

Let $h_n$ denote the number of non-negative integral solutions of the equation

$3x_1 + 4x_2 + 2x_3 + 5x_4  =  n$

Find the generating function $g(x)$ for $h_0,h_1,h_2,h_3 ... h_n$

Comments

I think this question should be some what same as - "Distributing n toffees to m children such that no one should get less than 2 toffees and more than 5 toffees"

Assumptions - m<n other wise the number of children will be more than Toffees which will contradict the statement our "no one should get less than 2 toffees".

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Answer 1

For number of non negative integral solution is n-11Cn-14

Comments

How is this solution obtained? Thanks

Answer 2

I got it partially as finding coefficient $x^{n}$ using Newton binomial expansion is not a human task, computers should be used.
The expansion looks like this:-

$x_{1}\in {(0,3,6,9,12.....)} \\ x_{2}\in {(0,4,8,12,16....)} \\ x_{3}\in {(0,2,4,6,8,.....)} \\ x_{4}\in {(0,5,10,15,20,.....)}$

$\begin{align*} &= [x^{n}](1+x^{3}+x^{6}........)(1+x^{4}+x^{8}+....)(1+x^{2}+x^{4}+....)(1+x^{5}+x^{10}+....)\\ &=[x^{n}]\frac{1}{1-x^{3}}.\frac{1}{1-x^{4}}.\frac{1}{1-x^{2}}.\frac{1}{1-x^{5}} \\ &=[x^{n}]\frac{1}{(1+x)^{2}}.\frac{1}{(1-x)^{2}}.\frac{1}{1+x^{2}}.\frac{1}{1-x^{3}}.\frac{1}{1-x^{5}} \\ &= [x^{n}]? \end{align*}$