# General formula for subtraction of binomial coefficients

I was playing around with the concept of subtraction of binomial expresions such as $\binom{n+k}{2}-\binom{n}{2}$, $\binom{n+k}{3}-\binom{n}{3}$, etc...

I was wondering if there was a known general formula for the expression $\binom{n+k}{m}-\binom{n}{m}$, with $n>k>m$. Ideally, I'm looking for general formulas which only make use of binomial coefficients.

I've figured out the first couple of terms, but I can't seem to find the general formula:

$\binom{n+k}{2}-\binom{n}{2}=\binom{k}{2}+\binom{k}{1}\binom{n}{1}$

$\binom{n+k}{3}-\binom{n}{3}=\binom{k}{3}+\binom{k}{2}\binom{n}{1}+\binom{k}{1}\binom{n}{2}$

$\binom{n+k}{4}-\binom{n}{4}=\binom{k}{4}+\binom{k}{3}\binom{n}{1}+\binom{k}{1}\binom{n}{3}+\binom{k}{2}\binom{n}{2}-\frac{11}{24}\binom{k}{1}\binom{n}{1}$*

*I am not completely sure that the math behind this one is totally correct. It does seem to break an otherwise natural trend.

• Notice that the formula for $\binom{n}{r}$ is $\frac{n!}{r!(n-r)!}$ – Franklin Pezzuti Dyer May 12 '17 at 19:26
Note that $${n+j \choose m} - {n+j-1 \choose m} = {n+j-1 \choose m-1}$$ so that \eqalign{{n + k \choose m} - {n \choose m} &= \sum_{j=1}^k \left({n+j \choose m} - {n+j-1 \choose m} \right)\cr &= \sum_{j=1}^k {n+j-1 \choose m-1}}
$${n+k \choose m} = \sum_{j=0}^m {k \choose m-j} {n \choose j}$$