Show that $ \frac{\sum_{n=1}^{49} n(52-n)(51-n)(50-n)}{13 \cdot 51 \cdot 50 \cdot 49} = 10.6$

Question

I have a hard time showing that \begin{align*} \frac{\sum_{n=1}^{49} n(52-n)(51-n)(50-n)}{13 \cdot 51 \cdot 50 \cdot 49} &= \frac{1}{13 \cdot 51 \cdot 50 \cdot 49} \left[ (1 \cdot 51 \cdot50\cdot 49) +(2 \cdot 50 \cdot 49 \cdot 48) + \ldots + (49 \cdot 3 \cdot 2 \cdot 1) \right]\\ &=10.6 \end{align*} I am not sure how to deal with/simplify the sum of the running products $(52-n)(51-n)(50-n)$.

Any help would be greatly appreciated.

Thank you.

Answer 1

Solution: $\begin{align*} \frac{\sum_{n=1}^{49} n(52-n)(51-n)(50-n)}{13 \cdot 51 \cdot 50 \cdot 49} &= \frac{1}{13 \cdot 51 \cdot 50 \cdot 49} \left[ \sum_{n=1}^{49}(132600n-7802n ^2+153n^3-n^4) \right]\\ \end{align*}$

As,

$\begin{align*} {\sum_{k=1}^{n} k=\frac{ n(n+1)}{2} }\end{align*}$

$\begin{align*} {\sum_{k=1}^{n} k^2=\frac{ n(n+1)(2n+1)}{6} }\end{align*}$

$\begin{align*} {\sum_{k=1}^{n} k^3=\frac{ n^2(n+1)^2}{4} }\end{align*}$

$\begin{align*} {\sum_{k=1}^{n} k^4=\frac{ n(n+1)(2n+1)(3n^2+3n-1)}{30} }\end{align*}$

Therefore,

$\begin{align*} \frac{1}{13 \cdot 51 \cdot 50 \cdot 49} \left[ \sum_{n=1}^{49}(132600n-7802n ^2+153n^3-n^4) \right]=\frac{1}{13 \cdot 51 \cdot 50 \cdot 49} \left[ 17218110 \right]=10.6\\ \end{align*}$

Answer 2

By reversing the order of summation $n \mapsto 52-n$, the desired identity is equivalent to $$\sum_{n=3}^{51} \frac{(52-n)n(n-1)(n-2)}{13 \cdot 51\cdot 50 \cdot 49}=\frac{53}{5}.$$ Now multiply both sides by $52 \cdot 51\cdot 50 \cdot 49/4!$ to obtain $$\sum_{n=3}^{51} \frac{(52-n)n(n-1)(n-2)}{3!}=\frac{53 \cdot 52 \cdot 51\cdot 50 \cdot 49}{5!},$$ which we can rewrite as $$\sum_{n=3}^{51} \binom{n}{3}\binom{52-n}{1}=\binom{53}{5}.$$ This last identity can be proved combinatorially by counting $5$-subsets of $\{1,\dots,53\}$ in two different ways. The RHS is clear. For the LHS, condition on the fourth smallest value $n+1$. Then you must choose $3$ elements from $\{1,\dots,n\}$ and $1$ element from $\{n+2,\dots,53\}$.

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Alternatively, skip the first step of reversing the order of summation. Instead rewrite as $$\sum_{n=1}^{49} \binom{n}{1}\binom{52-n}{3}=\binom{53}{5},$$ which you can prove combinatorially by conditioning on the fourth largest (equivalently, second smallest) element $n+1$.