I know that the number of non-decreasing sequences of length $n$ and numbers in the sequence lying in the range $[l,r]$ is given by $$\binom{n+r-l}{n}$$
What is the formula to find the $$\sum_{n=1}^{N}{\binom{n+r-l}{n}}$$
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Are you asking for a simplification of $\displaystyle \sum_{n=1}^N {n+r-l \choose n}$ ? – Henry Apr 11 '15 at 10:41
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@Henry Yes, exactly – Nikhil Em Apr 11 '15 at 10:44
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As a hint, if you add $1$ you would then be looking at $$ \displaystyle \sum_{n=0}^N {n+r-l \choose n}$$ and it is worth looking at this for small values of $N$ and $r-l$ to spot a pattern related to Pascal's triangle.
It will give you the result to your original question of $$ \displaystyle {N+r-l+1 \choose N} -1.$$
Henry
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