Number of distinct quadruples of non-negative integers

Question

I needed to find the number of distinct quadruples $(x_i), i=1,2,3,4,5$ of non-negative integers such that $x_1 \geq 3, x_2 \geq 3, x_4 \geq 8, x_5 \leq 3$, such that $x_1+x_2+x_3+x_4+x_5 \leq 23$. My answer does not agree with the correction. My reasoning is : substitute variables $$y_1=x_1-3 \geq 0,y_2=x_2-3 \geq 0,y_3=x_3,y_4=x_4-8 \geq 0,y_5=x_5,$$ and find the number of tuples satisfying $$\begin{cases}y_1+y_2+y_3+y_4 \leq 9 & (y_5=0)\\y_1+y_2+y_3+y_4 \leq 8 & (y_5=1)\\y_1+y_2+y_3+y_4 \leq 7 & (y_5=2)\\y_1+y_2+y_3+y_4 \leq 6 & (y_5=3)\\\end{cases}.$$ However (and this is where I disagree/don't understand the correction), this is the exact same thing as the number of tuples satisfying $y_1+y_2+y_3+y_4 \leq 9$, as all the former ones are included in this latter one. Hence my solution is $\overset{9}{\underset{k=0}{\sum}}{k+3 \choose 3}=715,$ where it seems that I should have found $1750$. Did I go wrong somewhere ?

Answer 1

If you introduce a nonnegative integer slack variable $y_6$ so that you get four equalities in $y$, you obtain the correct count of $$\sum_{k=6}^9 \binom{k+4}{4}=1750$$

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Alternatively, count the number of nonnegative integer solutions to $$\sum_{j=1}^6 y_j = 9$$ and subtract the ones that have $y_5 \ge 4$: $$\sum_{j=1}^6 y_j = 5$$ This approach yields $$\binom{9+5}{5}-\binom{5+5}{5}=1750$$