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?
Answer
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
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