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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)?2012-05-30
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    what is equal your second result $(n)_2$, please2012-05-30
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    $(n)_a$ 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 $n^{(a)}$)2012-05-30
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    Possibly you want to include ordered pairs $(k,k)$, in which case the answer is $n^2$.2012-05-30
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    @Shaktal $(n)_a$ is $\binom{n}{a} * a!$ like said in the ncmath's answer?2012-05-31

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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?2012-05-30
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    @ca11111: No, just the opposite: there are more *ordered* than *unordered*.2012-05-31
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    ah right (a,b) and (b,a) is the same unordered pair2012-05-31