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In how many ways can we get a sum greater than $x$ for $n$ distinct integers between $1$ and $x-1$, both inclusive?

For example

  • For $x = 5$ and $n = 3$, the required combinations are $(1,2,3)$, $(1,2,4)$, $(1,3,4)$ and $(2,3,4)$ i.e. totally 4 ways.
  • Similarly for $x = 6$ and $n = 3$, the required number of ways are $9$ which are $(1,2,4)$, $(1,2,5)$, $(1,3,4)$, $(1,3,5)$, $(1,4,5)$, $(2,3,4)$, $(2,3,5)$, $(2,4,5)$ and $(3,4,5)$.

Please help me answer this question and provide a detailed explanation if possible.

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