Combinatorics: how many different triples of cards can be formed let's say i have a hand of cards.
the number of cards i have in my hand is x > 3.
how many unordered different triples of cards can i form with the cards in my hand?
example:
i have the following cards in my hand: A B C D
i could form 
A B C
A B D
A C D
B C D
4 different triples can be formed.
i am looking for a formular that gives me the number of triples depending on x.

First let's look at the case where order matters. The number of possible first cards is $ x $, the number of possible second cards is $ x - 1 $ (since the first one is no longer available), and the number of possible third cards is $ x-2 $. The total number of three-card hands, where order matters, is then $ x(x-1)(x-2)$. However, given any three distinct cards, there are $ 3! = 6 $ different ways to order them all, e.g. 
ABC
ACB
BAC
BCA
CAB
CBA
Hence our value of $ x(x-1)(x-2) $ contains 6 copies of every possible three-card hand (without ordering). Thus the value you are looking for is $ x(x-1)(x-2)/6 $. Note that, in general, the number of ways of choosing $ m $ objects out of a collection of $ n $ objects, without order, is determined by the binomial coefficient:
$$ {n \choose m} = \frac{n!}{(n-m)! m!}, $$
which can be derived by the same reasoning process used above.
