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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





4 different triples can be formed.

i am looking for a formular that gives me the number of triples depending on x.

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up vote 3 down vote accepted

You have $x$ possibilities to choose the first card, $x-1$ for choosing the second one and $x-2$ for choosing the third one. If you multiply these numbers, you get $x(x-1)(x-2)$. However, as you look for unoredered triples, you have to divide this by the number of all possible ordering of 3 cards, which is $3!$ and you get $$\frac{x(x-1)(x-2)}6.$$

Nothe that this is the same as $\binom x3$, what is exactly what you would expect. See

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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.


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.

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There is a formula and a name. It's called a combination. You should also check out permutations. The notation is $n \choose 3$, or in this case $4 \choose 3$.

In general, the $n \choose k$ notation means $\dfrac{n!}{(n-k)! k!}$. So ${4 \choose 3} = \frac{4 \cdot 3!}{1 \cdot c!} = 4$.

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