For each $m \geq 0$, extract the coefficient of $x^m$ from both sides of this power series identity, $c \geq 0, t \geq 1$:
$$\frac{1}{(1 - x)^t} = \frac{(1 - x)^c}{(1 - x)^{t + c}}.$$
What binomial coefficient identity results? What does it mean in terms of counting combinatorial objects?
I tried to rewrite the above equation using binomial series. I got $$\sum_{n = 0}^\infty {n + t - 1 \choose t - 1} x^n = \sum_{k = 0}^c {c \choose k} (-1)^k(x)^k \cdot \sum_{n = 0}^\infty {n + t + c - 1 \choose t + c - 1} x^n$$
But I don't know how to extract a separate coefficient of $x^m$ from there. Any hints or suggestions?
The product of two generating functions $\sum_m a_m x^m$ and $\sum_m b_m x^m$ is the generating function of the convolution $\sum_{k=0}^m a_k b_{m-k}$, so \begin{align} \frac{(1-x)^c}{(1-x)^{t+c}}&= \left(\sum_{m=0}^\infty \binom{c}{m}(-x)^m\right)\left(\sum_{m=0}^\infty \binom{m+t+c-1}{t+c-1}x^m\right)\\ &= \sum_{m=0}^\infty \sum_{k=0}^m \binom{c}{k}(-1)^k \binom{m-k+t+c-1}{t+c-1} x^m. \end{align} Now extracting the coefficient of $x^m$ yields $$\binom{m+t-1}{t-1}=\sum_{k=0}^m \binom{c}{k}(-1)^k \binom{m-k+t+c-1}{t+c-1}.$$