combinatoric question - balls arrangement In how many ways can you arrange 200 balls into 40 cells, such that the sum of balls in cells 1-20 is greater (not equal) the sum of balls in cells 21-40?  
So, the number of posibilities can be described by:  
$$\sum\limits_{k = 0}^{99} {\left( {\begin{array}{*{20}{c}}
   {k + 20 - 1}  \\
   {20 - 1}  \\
\end{array}} \right)}$$
I was aked not to use $\sum$.
Is there another form to express it? 
 A: You can use the combinatorial identity $\sum_{k=0}^n\binom{k}i=\binom{n+1}{i+1}$
So $\sum_{k=0}^{99}\binom{k+20-1}{20-1}=\sum_{k=0}^{118}\binom{k}{19}-\sum_{k=0}^{18}\binom{k}{19}=\sum_{k=0}^{118}\binom{k}{19}=\binom{119}{20}$
$=24551856075980529765105$

To understand the combinatorial identity assume you want to choose k integers from the set $\{1,2,3...n\}$ Now look at the combinations of $k+1$ elements picked from $\{1,2,3...n,n+1\}$ For a given $m$ how many combinations have m as the largest number? $\binom{m-1}{k}$ Since there are k left to pick and they must be between $m-1$ and $1$. Clearly all combinations of $k+1$ elements have a largest element between $1$ and $m+1$. So the number of combinations of $k+1$ elements picked from $m+1$ elements is $\sum_{k=0}^m\binom{m+1}{k}$
A: $$24551856075980529765105$$
by wolfram alpha. This is one of the expressions without using $\sum$. 
A: Just for the record, the truly correct answer is
$${1\over2}\left({200+40-1\choose40-1}-{100+20-1\choose20-1}^2\right)$$
The ${200+40-1\choose40-1}$ counts the number of ways you can distribute the balls without any restriction.  The $-{100+20-1\choose20-1}^2$ subtracts off the number of arrangements with an equal number in cells 1-20 and 21-40.  The factor $1\over2$ takes care of the symmetry between the two groups of cells.  (For every arrangement with more in one group, there's a corresponding arrangement with more in the other.)  You can simplify this to
$${1\over2}\left({239\choose39}-{119\choose19}^2\right)$$
No doubt Mathematica or Whateverica can compute the exact value in decimal form; it'll have a lot of digits.
