Integer solutions to this problem

So I have this question, and I know that I need to start off with downgrading $782$ to $762$ (to account for $x_4$ and $x_5$ being equal to $10$).

$x_1 +x_2+x_3+x_4+x_5 ≤ 782$

where $x_1,x_2 > 0$, $x_3 ≥ 0$

$x_4,x_5 \geq 10$.

How would I go about solving this using the combination formula after deflating $782$ to $762$?

Let $x_6 = 782 - (x_1 + x_2 + x_3 + x_4 + x_5)$. Then the inequality

$$x_1 + x_2 + x_3 + x_4 + x_5 \leq 782$$

becomes the equation

$$x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 782$$

where $x_1, x_2$ are positive integers, $x_3, x_6$ are non-negative integers, and $x_4, x_5 \geq 10$. Let \begin{align*} y_1 & = x_1 - 1\\ y_2 & = x_1 - 1\\ y_3 & = x_3\\ y_4 & = x_4 - 10\\ y_5 & = x_5 - 10\\ y_6 & = x_6 \end{align*} Then each $y_i$, $1 \leq i \leq 6$, is a non-negative integer, and \begin{align*} y_1 + 1 + y_2 + 1 + y_3 + y_4 + 10 + y_5 + 10 + y_6 & = 782\\ y_1 + y_2 + y_3 + y_4 + y_5 + y_6 & = 760 \end{align*} The number of solutions of this equation in the non-negative integers is the number of ways five addition signs can be placed in a list of $760$ ones, which is $$\binom{760 + 5}{5} = \binom{765}{5}$$

$y_1=x_1+1, y_2=x_2+1,y_3=x_3, y_4=x_4+10,y_5=x_5+10, y_6 \geq 0 \to (y_1+1)+(y_2+1)+y_3+(y_4+10)+(y_5+10)+y_6 = 782 \to y_1+y_2+y_3+y_4+y_5+y_6=762$. The number of solutions is:

$\binom{762+6-1}{6-1} = \binom{767}{5}$

• Be careful, you have $$y_1 + 1 + y_2 + 1 + y_3 + y_4 + 10 + y_5 + 10 = 762$$ which implies $$y_1 + y_2 + y_3 + y_4 + y_5 + y_6 = 760$$ Also, $y_1 = x_1 - 1$, $y_2 = x_2 - 1$, $y_4 = x_4 - 10$, and $y_5 = x_5 - 10$. – N. F. Taussig Jan 14 '15 at 23:04

Just as you already know, we first introduce the variables $y_1=x_1$, $y_2=x_2$, $y_3=x_3+1$, $y_4=x_4-9$ and $y_5=x_5-9$. Then your problem is equivalent to finding the number of integer solutions for the equation $$y_1+y_2+y_3+y_4+y_5\leq 765,$$ where $y_i\geq 1$.

The usual way of solving this kind of problem is the following: Suppose we have $765$ points in a line: $\bullet-\bullet-\cdots-\bullet$. You can think of a solution $y_1,\ldots,y_5$ as the following: take the set $A_1$ of the first $y_1$ points, then the set $A_2$ of the next $y_2$ points, $A_3$ the next $y_3$ points, same for $A_4$ and $A_5$: $$\underbrace{\bullet-\bullet-\cdots-\bullet}_{A_1}-\underbrace{\bullet-\cdots-\bullet}_{A_2}\cdots\underbrace{\bullet-\cdots-\bullet}_{A_5}-\bullet-\cdots-\bullet$$ so what really matters is the point at which each of the sets $A_1$, $A_2$, $A_3$, $A_4$ and $A_5$ finishes, that is, we have to calculate the number of choices of $5$ points among all those $765$ (and the order does not matter), which is $\binom{765}{5}$.

Another way to see this problem is dealing with each case separately:

• before that we will modify the variable: $y_4 = x_4 -10 \geq 0$ , $y_5 = x_5 -10 \geq 0$ , $y_1=x_1 -1 \geq 0$, and $y_2 = x_2 -1 \geq 0$ our new system is $x_1 +x_2+x_3+y_4+y_5 \leq 760$

• $y_1 +y_2+x_3+y_4+y_5 = 0\$ ... , $\ y_1 +y_2+x_3+y_4+y_5 = k$ ... , $\ y_1 +y_2+x_3+y_4+y_5 = 760$ each case is disjoint the number of solutions is:

$N=\sum_{k=0}^{760} \binom{k-1+5}{5}$

• Be careful: $x_1, x_2 > 0$. – N. F. Taussig Jan 14 '15 at 23:01
• ops thats right, I need to modify the answer. Thank you. – SomeOne Jan 14 '15 at 23:02