Number of natural solutions to $x_1 + x_2 + x_3 + 2x_4 + x_5 = 72$

Question

What are the number of natural solutions to $$x_1 + x_2 + x_3 + 2x_4 + x_5 = 72$$ where $x_1 \ge 2, x_2, x_3 \ge 1, x_4, x_5 \ge 0$?

I understand how to do it if it wasn't "$2x_4$", hence if the coefficient of $x_4$ was $1$, , then the answer will be $C(72,4) \ldots \ $, but given $2x_4$, I don't know how to solve the question.

Answer 1

Move $2x_4$ to the other side and solve a separate $4$-variable problem for each possible value of $x_4$. In other words, you’re counting non-negative solutions to

$$x_1+x_2+x_3+x_5=68-2k$$

for $k=0,\ldots,34$, and you get

$$\sum_{k=0}^{34}\binom{71-2k}3=\sum_{k=1}^{35}\binom{2k+1}3\;.$$

This actually isn’t quite as nasty as it may look. If we calculate the first few values of $$a_n=\sum_{k=1}^n\binom{2k+1}3\;,$$ we get $a_1=1$, $a_2=10$, $a_3=35$, and $a_4=84$, with first differences $9$, $25$, and $49$. That suggests that we’re looking at sums of odd squares, i.e., that

$$\begin{align*} \sum_{k=1}^{n+1}\binom{2k}3&=\sum_{k=1}^n(2k-1)^2\\ &=4\sum_{k=1}^nk^2-4\sum_{k=1}^nk+\sum_{k=1}^n1\\ &=\frac23n(n+1)(2n+1)-2n(n+1)+n\\ &=\frac13n(4n^2-1)\;. \end{align*}$$

This can be straightforwardly proved by induction on $n$.

Answer 2

EDIT: Thanks @Brian M. Scott for enlightening me. I was confusing two different things.

First up, the different conditions

$$x_1 \ge 2; x_2, x_3 \ge 1; x_4, x_5 \ge 0$$

are rather inconvenient. Let us change the problem to finding the solutions of

$$y_1 + y_2 + y_3 + 2y_4 + y_5 = 68$$ with $y_i \ge 0\ \forall_{i \le 5}$

Where we subtracted 2 from $x_1$ and subtracted 1 from each $x_2$ and $x_3$. But then again it is quite inconvenient to have the 2 factor in the middle of the expression. Let us move it to the beginning, shall we? Renaming the variables, we want to solve

$$2x_1 + x_2 + x_3 + x_4 + x_5 = 68, x_i \ge 0$$

right? If you fix the value of $x_1 = 2k$, then you want to solve

$$x_2 + x_3 + x_4 + x_5 = 68 - 2k, x_i \ge 0$$

right? But that is an easily solvable problem. Since the number of solutions of that equation is given by $C(68 - 2k, 4)$, your answer is just summing up through all values of $68 - 2k$:

$$\sum_{i = 0}^{34} C(68 - 2k, 4)$$

Answer 3

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What are the number of integer solutions to \begin{equation} x_{1} + x_{2} + x_{3} + 2x_{4} + x_{5} = 72\ ?\quad\mbox{where}\quad x_{1} \geq 2\,;\quad\ x_{2}\,,\ x_{3} \geq 1\,;\quad\ x_{4}\,,\ x_{5} \geq 0 \label{1}\tag{1} \end{equation}

That is given by \begin{align} &\sum_{x_{1} = 2}^{\infty}\ \sum_{x_{2} = 1}^{\infty}\ \sum_{x_{3} = 1}^{\infty}\ \sum_{x_{4} = 0}^{\infty}\ \sum_{x_{5} = 0}^{\infty} \bracks{x_{1} + x_{2} + x_{3} + 2x_{4} + x_{5} = 72} \\[5mm] = &\ \sum_{x_{1} = 0}^{\infty}\ \sum_{x_{2} = 0}^{\infty}\ \sum_{x_{3} = 0}^{\infty}\ \sum_{x_{4} = 0}^{\infty}\ \sum_{x_{5} = 0}^{\infty} \bracks{x_{1} + x_{2} + x_{3} + 2x_{4} + x_{5} = 68} = \bracks{z^{68}}\bracks{1 \over \pars{1 - z}^{5}\pars{1 + z}} \\[5mm] = &\ \bbox[#ffe,15px,border:2px dotted navy]{\ds{528,990}} \end{align}