Simplify a Combinatorial Sum $\sum_{k=0}^\infty {a\choose k}{b\choose c-k}{d-k\choose e}$

Question

Is there a way to simplify $$\sum_{k=0}^\infty {a\choose k}{b\choose c-k}{d-k\choose e}$$ where $a,b,c,d,e$ are natural numbers?

In particular, I would like to see the case for $a=45, b=3,c=4,d=48,e=4$.

Answer 1

The series can be cast into hypergeometric form and is given by \begin{align} \sum_{k=0}^\infty \binom{a}{k} \binom{b}{c-k} \binom{d-k}{e} = \binom{b}{c} \binom{d}{e} \, {}_{3}F_{2}(-a, -c, e-d; -d, b-c+1; 1) \end{align} Further reduction may be possible depending on the values of $\{a,b,c,d,e\}$.

Proof of result \begin{align} S &= \sum_{k=0}^\infty \binom{a}{k} \binom{b}{c-k} \binom{d-k}{e} \\ &= \sum_{k=0}^{\infty} \frac{b! \, \Gamma(a+1) \Gamma(d+1-k)}{k! \, \Gamma(a+1-k) \Gamma(c+1-k) \Gamma(b-c+1+k) \, e! \, \Gamma(d-e+1-k)} \\ &= \frac{b! \, d!}{e! \, c! \, (b-c)!} \, \sum_{k=0}^{\infty} \frac{(-a)_{k} (-c)_{k} (e-d)_{k}}{k! \, (-d)_{k} (b-c+1)_{k}} \\ &= \binom{b}{c} \binom{d}{e} \, {}_{3}F_{2}(-a, -c, e-d; -d, b-c+1; 1). \end{align} It is to be noted that $b-c+1 \neq 0,-1,-2, \cdots$ and $d \leq 0$.

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As given by the example values $a=45, b=3,c=4,d=48,e=4$ the series is best handled in a straight forward manor. \begin{align} \sum_{k=0}^\infty \binom{45}{k} \binom{3}{4-k} \binom{48-k}{4} &= \sum_{k=1}^{4} \binom{45}{k} \binom{3}{4-k} \binom{48-k}{4} \\ &= 45 \, \binom{47}{4} + 3 \cdot 45 \cdot 22 \, \binom{46}{4} + 3 \, \binom{45}{3} \binom{45}{4} + \binom{44}{4} \binom{45}{4} \\ &= 27,061, 623, 270. \end{align}