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All the question is in the title:

How many 2-tuple can I form from the elements of a n-tuple
Supposing all elements are differents also (eg: (A,C,D,I,X,Y))

is it simply $\binom{n}{2}$ ?

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Is it not simply the number of valid combinations of pairs you can form (if the order is unimportant), i.e. $\binom{n}{2}$, or the number of permutations of the set into 2-tuples: $(n)_2$ (if order is important)? – Shaktal May 30 '12 at 14:36
what is equal your second result $(n)_2$, please – caub May 30 '12 at 14:46
$(n)_a$ is defined as $n(n-1)(n-2)...(n-a+1)$, it's called the pochhammer symbol (also known as the falling factorial when it's denoted as $n^{(a)}$) – Shaktal May 30 '12 at 15:00
Possibly you want to include ordered pairs $(k,k)$, in which case the answer is $n^2$. – André Nicolas May 30 '12 at 15:20
@Shaktal $(n)_a$ is $\binom{n}{a} * a!$ like said in the ncmath's answer? – caub May 31 '12 at 11:23
up vote 1 down vote accepted

If you are looking for subsets of size 2, it's $n\choose 2$. If you want ordered pairs of size 2, there are twice as many of these, so it's $n(n-1) = 2{n\choose 2}$.

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there should be more unordered than ordered? – caub May 30 '12 at 15:07
@ca11111: No, just the opposite: there are more ordered than unordered. – Brian M. Scott May 31 '12 at 0:26
ah right (a,b) and (b,a) is the same unordered pair – caub May 31 '12 at 11:20

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