If yes, how to arrive at that formula?

What are the basic steps?

Ans :

For m terms to the nth power, m >= 1 and n >= 0,

u(*,0) = 1

u(0,n+1) = 0

u(1,n) = 1

u(2,n) = n+1

u(m,1) = m

u(m,2) = m*(m+1)/2

u(m+1,n+1) = u(m+1,n) + u(m,n+1)

That last equation looks likes the recurrence relation for Pascal's triangle, so there's a close relationship there.

u(m,n) = C(m+n-1,n) where C(x,y) = x!/(y! * (x-y)!)

u(m,n) = (m+n-1)!/(n! * (m+n-1-n)!) = (m+n-1)!/(n! * (m-1)!)

Let's check it. Remember that 0! = 1

u(1,1) = 1 1!/(1!*0!) = 1

u(2,1) = 2 2!/(1!*1!) = 2

u(n,1) = n n!/(1! * (n-1)!) = n

u(1,2) = 1 2!/(2! * 0!) = 1

u(2,2) = 3 3!/(2! * 1!) = 3

u(n,2) = n*(n+1)/2 (n+1)!/(2! * (n-1)!) = n*(n+1)/2

u(2,n) = n+1 (n+1)!/(n! * 1!) = n+1

So (a+b+c+...)^n with m variables has (m+n-1)!/(n! * (m-1)!) distinct terms

Ans 2 :

http://mathforum.org/library/drmath/view/51601.html

Ans 3 :

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