This is a problem for a practice test my professor gave me.
$$\text{How many integer solutions are there to } x_1+x_2+x_3+x_4 \leq 50 \\ \text{with } x_i \geq 2 \text{ for all } i = 1,2,3,4 \text{ and } x_1,x_2 \leq 7 \text{?}$$
This is how I approached the problem, using generating functions:
$\text{Same as}$
$$x_1+x_2+x_3+x_4+x_5 = 50, \space x_5 \geq 0 $$
$\text{Find the coefficient of } x^{50} \text{ in}$
$$(x^2 + x^3 + x^4 + \dotso)^2 (x^2 + x^3 + x^4 + \dotso + x^7)^2 (1 + x + x^2 + x^3 + \dotso)$$
After some factoring, we'll have:
$$x^8(1+x^2+x^3+ \dotso + x^5)^2 (1+x+x^2+\dotso)^3$$
This is the same as:
$\text{Find the coefficient of } x^{42} \text{ in}$
$$(1+x^2+x^3+ \dotso + x^5)^2 (1+x+x^2+\dotso)^3$$
To simplify further:
$$(1-x^6)^2 \frac{1}{(1-x)^3}$$
So, this is where I'm confused. I was using the formula which is based off this answer on Math.SE, but I don't get the correct answer. According to my professor, the correct answer is:
$$\dbinom{30+5-1}{30} – 2\dbinom{36+5-1}{36} + \dbinom{42+5-1}{42} = 26,781$$
What I end up doing mirrors that of the linked question on Math.SE:
$$(1-x^6)^2 = 1-2x^6+x^{12} \\
\\
(1-2x^6+x^{12}) \frac{1}{(1-x)^3}$$
Using the formula from the linked question:
$\text{We do this three times, for } k=0, k=6, \text{ and } k=12$. The result is
$$(1-2x^6+x^{12})\frac{1}{(1-x)^3}={m-0+2 \choose 2}- 2{m-6+2 \choose 2} + {m-12+2 \choose 2}
\\
\
\\m = 42 \\
\
\\={42-0+2 \choose 2}- 2{42-6+2 \choose 2} + {42-12+2 \choose 2} = 36$$
As you can see, my answer differs greatly from what my professor said was correct. I don't understand why this formula I used doesn't work; I've used it for lots of other problems of this same type, and I calculated the correct number; for this one though, it doesn't seem to be working.
What am I doing wrong?
Best Answer
The actual generating function should be $$\frac{(1-x^6)^2}{(1-x)^5}$$ When you said, "To further simplify...," you forgot that $$1+x+x^2+x^3+x^4+x^5=\frac{1-x^6}{1-x},$$ not simply $1-x^6$.