How do you calculate %xHKO?

I asked my friend about this, heres the chat log (I do not know if this is correct or not)

Me: I have a question in probablities
Me: assume you are in maple
Me: and there is a monster with x HP
Me: and your damage is L~H
Me: what is the probablity to kill it in n hits
Him: oh, I think I know how to solve it
Him: first you move from L~H to 0~K (K being H - L)
Him: then you solve the equation x1 + x2 + x3 + ... + xn = x (with each n being a natural number between 0 and K)
Him: now take this (1 + x + x^2 + x^3... + x^k)^n
Him: and you need to find the coefficient of x^k
Him: you can split it to (1-x^(k+1))^n * (1+x+x^2+x^3.......)^n
Him: then you need to open the finite part with Newton's Binomial
Him: for example if N is 2, you get: (1 - 2x^(k+1) + x^(2k+2)) * (1 + x ...)^n
Him: D(n, x) - 2D(n, x - k - 1) + D(n, x - 2k - 2)
Him: D(n, k) = C(n + k - 1, k)
Him: let me solve that
Him: I think thats the solution, if I didn't have any mistakes

Him: That circle in the beginning is supposed to be a capital Sigma
Him: This is supposed to give the number of successes, you should divide it with the number of possible options

Edit: he said he had a mistake and that this is the correct solution