# 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.

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.
You can still ask an expert for help

• Questions are typically answered in as fast as 30 minutes

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

Spielgutq1
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 $x\left(x-1\right)\left(x-2\right)$. 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
$\frac{x\left(x-1\right)\left(x-2\right)}{6}.$
Nothe that this is the same as $\left(\genfrac{}{}{0}{}{x}{3}\right)$, what is exactly what you would expect.
###### Did you like this example?
bergvolk0k
There is a formula and a name. It's called a combination. You should also check out permutations. The notation is $\left(\genfrac{}{}{0}{}{n}{3}\right)$, or in this case $\left(\genfrac{}{}{0}{}{4}{3}\right)$.
In general, the $\left(\genfrac{}{}{0}{}{n}{k}\right)$ notation means $\frac{n!}{\left(n-k\right)!k!}$. So $\left(\genfrac{}{}{0}{}{4}{3}\right)=\frac{4\cdot 3!}{1\cdot c!}=4$.