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I have the following sum

$$5\cdot4\cdot3+5\cdot4\cdot2+5\cdot4\cdot1+5\cdot3\cdot2+5\cdot3\cdot1+$$$$5\cdot2\cdot1+4\cdot3\cdot2+4\cdot3\cdot1+4\cdot2\cdot1+3\cdot2\cdot1$$

It is basically $5!$ divided by two of the numbers in the factorial. So

$$\frac{5!}{1\cdot2}+\frac{5!}{1\cdot3}+\frac{5!}{1\cdot4}+...+\frac{5!}{3\cdot5}+\frac{5!}{4\cdot5}$$

Is there a way to write this as a single summation?

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You can write it as a single sum as follows

$$\frac{5!}{1\cdot2}+\frac{5!}{1\cdot3}+\frac{5!}{1\cdot4}+...+\frac{5!}{3\cdot5}+\frac{5!}{4\cdot5}=5!\sum_{1\le i <j\le 5}\frac{1}{i\cdot j}.$$

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Just as an alternative:

$$\sum_{S\subset \{1,2,3,4,5\}}_{|S|=3}\prod_{i\in S}i$$

This generalizes to other values.

It is also the coefficient of $x^3$ in the polynomial $(1+x)(1+2x)(1+3x)(1+4x)(1+5x)$, for what it is worth.

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  • $\begingroup$ That is actually very helpful. Working with Bernoulli numbers and polynomials and there are a lot of combinatorial connections, so it might be adventageous to keep that generating polynomial in the back of my mind... Thanks! $\endgroup$ – Eleven-Eleven Dec 3 '14 at 1:15

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