>>8178555
Your question is incoherent. Try again later.

>>8178574
I mean binomial coefficient with some addition so it could be easy to calculate if you're not trolling.

>>8178584
Coefficient of x^n in x^K ( (x^6 - 1)/(x - 1) )^K where n is the number of interest.

>>8178595
Just to clarify.
I want number - n to represent with K summands, which are in range (1..6).
So I need to get binomial coefficient from your formula. But what x means there?

>>8178634
I don't think you understand the terms you're using or basic combinatorics.

x is not equal to some number. It represents the object being counted.

The binomial coefficient is a coefficient in the expansion of (x+y)^n. In the case of the binomial coefficient, you are counting how many instances of x^k y^(n-k) appear in the expansion, because this is equivalent to the number of ways to choose k objects out of n objects.

Similarly, the number of ways to add K positive integers less than or equal to 6 so they equal n is equal to the coefficient of x^n in the expansion of x^K ( (x^6 - 1)/(x - 1) )^K

>>8178748
Let me explain all the task I have.

Given N - dice, M - number I wish. Each dice has 6 edges (this is where 1..6 comes from)

I need to answer what probability of situation where I throw N dice and get M in result.
End of task.

To resolve this i need to divide positive results on all results that I can get throwing N dice.

To get positive results I want to calculate how much ways I can get M value.

I found that I can represent M as sum of element, even thou I can calculate number of ways I can represent M as a sum with N summands. And I can calculate this results as (n - 1 | k - 1) = (n - 1)! / ( (k-1)! * ((n-1) - (k - 1) )! ) It's here https://en.wikipedia.org/wiki/Composition_(combinatorics)
For this expression I can put my M and N and calculate.
But in this case you can get summand that is greater than 6, and this results shouldn't count.

I want to find way that I can really say how much ways there will be that sum == M, number of summands will be N, and each summand will be 1..6

>>8178813
>I want to find way that I can really say how much ways there will be that sum == M, number of summands will be N, and each summand will be 1..6
Yes, and I just gave you one. What is the issue?

>>8178813
Can you perhaps explain your task in a better formatted way? Your english is fucking shitty and you aren't stating your problem or goal simply enough or using proper vocabulary to describe it.

You may as well be shouting gibberish into a well.

>>8178843
Your answer are not clear for me. Can you represent it more clear?
Like I described it, for example:
(n | k) = n! / (k! * (n-k)!)
where ! is factorial.

>>8178846
I need to calculate probability of event that I get number M as sum when I throw N dice.

Simple enough?

>>8178868
You mean closed form? Fine:

sum from k=0 to floor((M-N)/6) of (-1)^k N (M-6k-1)! / ( k! (N-k)! (M-6k-N)! )

>>8178898
This works, thank you very much.

But can you explain it like if you explain it for a child. I still can't understand how you made it from expressions to sum of elements and how you create this expressions.

Tell if you don't mind to contact via skype or mail.