$\binom{54}{5}+\binom{49}{5}+\binom{44}{5}+\cdots+\binom{9}{5}$

Question

How to calculate the sum $$\binom{54}{5}+\binom{49}{5}+\binom{44}{5}+\cdots+\binom{9}{5}$$

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I wrote this as $$\sum_{r=2}^{11}\binom {5r-1}{5}$$ $$=\frac{1}{120}\sum_{r=2}^{11}(5r-1)(5r-2)(5r-3)(5r-4)(5r-5)$$ I'm stuck after this. Any help is greatly appreciated.

Answer 1

$$f:\mathbb Z \rightarrow \text{any commutative ring}$$$$\text {forward difference notation }\Delta f(x)=f(x+1)-f(x)$$ $$\sum_{x=a}^b \Delta f(x)=f(b+1)-f(a)$$ $$\text {falling power notation } x^{(k)}=x(x-1)…(x-k+1)$$ $$\text {fact: } x^{(k)}=(\Delta x^{(k+1)})/(k+1)$$ Your sum is $(1/5!)\sum_{x=9}^{54}x^{(5)}=(1/(5! \times 6))\sum_{x=9}^{54}\Delta x^{(6)}$ $$=(1/6!)(55^{(6)}-9^{(6)})$$ $$={55 \choose 6}-{9 \choose 6}$$

Answer 2

For a combinatorial interpretation of the sum, consider $6$-subsets of $\{1,\dots,55\}$ such that the largest element is a multiple of $5$. The sum conditions on the largest element $5r$; the remaining $5$ elements are then chosen from among $\{1,\dots,5r-1\}$.

Without the "multiple of $5$" restriction, this approach yields a combinatorial proof that $$\sum_{k=6}^{55} \binom{k-1}{5} = \binom{55}{6}.$$