Number of positive integer solutions of $x+y+z+w=26$ such that $x \lt y \lt z$

Question

Number of positive integer solutions of $x+y+z+w=26$ such that $x \lt y \lt z$

my try:

Let $y=x+p$ and $z=x+q$ where $p \ge 1$ and $q \ge 2$

Then we have

$$3x+p+q+w=26$$ $\implies$

$$p+q+w=26-3x$$

letting $q_1=q-1$ we get

$$p+q_1+w=25-3x$$ By stars and bars we get number of positive integer solutions as

$$\binom{25-3x-1}{3-1}=\binom{24-3x}{2}$$

But $x$ can take maximum value as $x=7$ since the last combination is $7,8,10,1$

So total solutions is

$$\sum_{x=1}^{7}\binom{24-3x}{2}=588$$

Is this correct approach?

Answer 1

Almost, but there is a mistake. When you set $y=x+p$ and $z=x+q$ with $p\ge 1$ and $q\ge 2$, this does not force $z > y$. For example, let $p=3$ and $q=2$.

A correct approach would be to set $$ \begin{array}{lll} y = x + p,&&p \ge 1.\\ z = x + p + q,&& q \ge 1.\quad \end{array} $$ Then you need to solve $$ 3x+2p+q+w = 26. $$

Answer 2

To verify your result, I would suggest an approach like this $$\sum_{x=1}^{7} \sum_{y=x+1}^{\lfloor\frac{26-x}{2}\rfloor} \sum_{z=y+1}^{26-x-y} (26-x-y-z) $$ who leads to 1914.

($26-x-y-z$: Number of possibilities for $w$)