Given $S=\{1,2,...,32\}$ and $T = \{(x_1, x_2,x_3,x_4)\in S^4|x_2 \geq x_1 +3, x_3 \geq x_2, x_4 \geq x_3 + 5\}$, find $|T|$

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Q: Let $S=\{1,2,...,32\}$ and $T = \{(x_1, x_2,x_3,x_4)\in S^4|x_2 \geq x_1 +3, x_3 \geq x_2, x_4 \geq x_3 + 5\}.$ Find $|T|$.

Answer provided(Using method for finding number of integer solutions):

Note: $x_4 \leq 32$.

Let:

$$y_1 = \hspace{10mm}x_1 \geq 1$$

$$y_2=x_2 -x_1 \geq 3$$

$$y_3 = x_3 -x_2 \geq 0$$

$$y_4 =x_4 -x_3 \geq 5$$

So, $y_1 +y_2 + y_3 +y_4 + y_5 = 32, $ for $y_1 \geq 1, y_2 \geq 3, y_3 \geq 0, y_4 \geq 5, y_5 \geq 0$.

Hence, we have

$$|T|= number \hspace{1mm} of \hspace{1mm}integer \hspace{1mm}solutions $$ $$=H_r^n$$ $$= {{r+n-1}\choose r}$$ $$= {{32-1-3-5+5-1}\choose{32-1-3-5}}$$ $$={27 \choose 23}$$ $$= {27 \choose 4}$$ $$\hspace{150mm}_{Q.E.D}$$

Question is, I do not understand why we have to include a $y_5 \geq 0$ when there are only 4 variables. Why doesn't $y_i$ for $i = 1,2,3,4$ suffice?

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$y_5$ is the "slack" variable; insisting it is nonnegative enforces that $x_4\le 32$.