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What is the number of non-empty subsets from the set $ (1,2,3,...,12)$ and such that the sum of the least element and the greatest element in the set is equal to $13$

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What have you tried already? Maybe you could work out what happens in the cases when your set is {6,7} or {5,6,7,8}. – JakeR Sep 21 '12 at 15:36
In standard usage one uses curly braces for sets and round brackets for tuples; thus $\{1,2,3,\ldots,12\}$ is a set and $(1,2,3,\ldots,12)$. A set does not become a different set if one lists the members in a different order or lists one of them more than once, but a tuple becomes a different tuple if one does that; that is the difference. – Michael Hardy Sep 21 '12 at 15:42

You need 1 and 12, or 2 and 11, or 3 and 10, or 4 and 9, or 5 and 8, or 6 and 7. And, once you have these, the number of elements still available depends on which group. The number of elements available would be the number between these, say $k$, so the number of subsets would be $2^k$. Therefore the number of subsets is

$$2^{10} + 2^8 + 2^6 + 2^4 + 2^2 + 2^0 = 1365$$

Just to be clear, here's one example. Say your least element is 4 and your greatest is 9. Then, your set must contain those 2 elements, but it also can contain any subset of the elements in between, $\{5, 6, 7, 8\}$. There are 4 elements in this, so there are $2^4$ possible subsets of this set.

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This can be computed in GAP via:

A:=Filtered(Combinations([1..12]),S->S<>[] and Maximum(S)+Minimum(S)=13);;

which gives 1365, matching Graphth's answer.

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