Vtyjkyuk

I am trying to solve a problem using permutation/combination but cannot figure out how to proceed.

Suppose the sum of three variables $x, y, z$ is $30$. If $xge2, yge0, zge-3$, how many integer solutions exist?

I understand that $2le xle33, 0le yle31, -3le zle28$. A simple simulation shows that there are $528$ solutions. However, I am unable to calculate this mathematically. I would like a hint so that I can try this on my own.

From
$$x+y+z=30 Rightarrow (x-2)+y+(z+3)=31 Rightarrow x_1+y+z_1=31\
x_1geq0,ygeq0,z_1geq0$$
which has $binom332=528$ integer solutions.

With more details, this is the generating function:

$$(1+x+x^2+...+x^k+...)^3=frac1(1-x)^3=
frac12left(frac11-xright)^''=\
frac12left(sumlimits_k=0x^kright)^''=
sumlimits_k=2frack(k-1)2x^k-2$$
and the coefficient of $x^31$ is the answer.

Set $a=x-2$, $b=y$ and $c=z+3$.

The problem is the same as finding the number of sums $a+b+c=31$ where $a,b,c$ are nonnegative integers.

Now apply stars and bars.

I started by fixing x=2, and counting all the solutions in y and z. For x = 2, we need $y+z=28$. Now, fixing y determines z. For y=0, z=28; y=1, z=27; ...; y=31, z = -3. So, we can count 32 solutions when x=2.

Then, I fixed x=3 and counted all the solutions in y and z again. A pattern emerges quickly.

Another hint:

$sum_i=0^32$ i = 528$

We wish to find the number of solutions of the equation
$$x + y + z = 30 tag1$$
in the integers subject to the restrictions that $x geq 2$, $y geq 0$, and $z geq -3$.

Let $x' = x - 2$, $y' = y$, and $z' = z + 3$. Then $x', y', z'$ are nonnegative integers. Substituting $x' + 2$ for $x$, $y'$ for $y$, and $z' - 3$ for $z$ in equation 1 yields
beginalign*
x' - 2 + y' + z' + 3 & = 30\
x' + y' + z' & = 31 tag2
endalign*
Equation 2 is an equation in the nonnegative integers. A particular solution of equation 2 corresponds to the placement of two addition signs in a row of $31$ ones. For instance,
$$1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1 1 +$$
corresponds to the solution $x_1 = 21$, $x_2 = 10$, $x_3 = 0$, while
$$1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1 1 1 + 1 1 1 1 1 1 1 1 1 1 1 1 1$$
corresponds to the solution $x_1 = 7$, $x_2 = 11$, $x_3 = 13$. The number of solutions of equation 2 is the number of ways we can place two addition signs in a row of $31$ ones, which is equal to the number of ways we can select which $2$ of the $33$ positions required for $31$ ones and $2$ addition signs will be filled with addition signs.

$$binom31 + 22 = binom332 = frac33!2!31! = frac33 cdot 32 cdot 31!2 cdot 1 cdot 31! = frac33 cdot 322 = 33 cdot 16 = 528$$