How many solutions exists for this equation?

Question

$$x_1 + x_2 + x_3 + x_4 = 28$$ I tried to solve it with generating functions. Is it correct to get to the form of

$${(1 + x + {x^2} + {x^3} + ....)^4}$$

and this equals to:

$${(1 - x)^{ - 4}}$$

and then solve it with binomial expansion ? Is this correct ?

every question similar to "how many solutions are axsist to this equation?" can be solved this method?

I know there're a lot of methods and manipulations for this kind of problem. Can you confirm it's the right path for this kind of problems?

It's like distribute 28 things into 4 objects. This is why I used generating functions.

Answer 1

I will assume you are asking for how many solutions with $x_i\in\mathbb{N}$ for each $i$ (where $\mathbb{N}=\{0,1,2,\dots\}$). If you are looking for real solutions, there are clearly infinitely many.

The number of integral solutions to $x_1+x_2+\dots+x_k=n$ with each $x_i\geq 0$ is given as $$\binom{n+k-1}{k-1}$$

This can be seen via stars&bars.

Answer 2

Yes, the binomial expansion of $(1-x)^{-4}$ will work.

No, many other type of questions like this can not be solved by this method.

Here is a useful expansion:

$(1-x)^{-n} =\sum_{k=0}^{\infty} \binom{-n}{k}(-x)^k =\sum_{k=0}^{\infty} (-1)^k\binom{-n}{k}x^k $.

And,

$\begin{array}\\ \binom{-n}{k} &=\frac{(-n)(-n-1)(-n-2)...(-n-k+1)}{k!}\\ &=\frac{(-1)^k(n)(n+1)(n+2)...(n+k-1)}{k!}\\ &=(-1)^k\frac{(n-1)!(n)(n+1)(n+2)...(n+k-1)}{k!(n-1)!}\\ &=(-1)^k\binom{n+k-1}{k}\\ \end{array} $

Therefore, noting that the two $(-1)^k$s cancel each other out, $(1-x)^{-n} =\sum_{k=0}^{\infty} \binom{n+k-1}{k}x^k $.