# Probability - Combinatorics

## Main Question or Discussion Point

Let's say we have a deck with 40 cards. There are two of each for each of the 4 suits: 10, Jack, Queen, King, and Ace. Each hand consists of 10 cards.

Given that each pair is technically the same (one 10 of hearts is not distinguishable from the other 10 of hearts), how would one calculate the number of possible hands?

You can't just do (40 choose 10) given my last statement.

I know that if I were just doing the number of possible orderings of the deck, I could do (40!)/(2!^20), but I don't think I can apply the same method to when I'm choosing a hand of 10.

Related Set Theory, Logic, Probability, Statistics News on Phys.org
chiro
Hey brojesus111 and welcome to the forums.

One thing you can do is to group all the same numbers together and then use combinations to get the number of combinations of different numbers in a hand and then for each of these, look at these groups and see how many possibilities you have for each number.

So as an example, lets say you have three different types of numbers: you take a combination of 10C3 but once you have done this, you look at the variation within these possibilities: for example if you restrict only three types of numbers then you can have say the following internal number configurations: (1,1,3) and (1,2,2).

This will not distinguish what the actual card is: only it's face value.

Hey brojesus111 and welcome to the forums.

One thing you can do is to group all the same numbers together and then use combinations to get the number of combinations of different numbers in a hand and then for each of these, look at these groups and see how many possibilities you have for each number.

So as an example, lets say you have three different types of numbers: you take a combination of 10C3 but once you have done this, you look at the variation within these possibilities: for example if you restrict only three types of numbers then you can have say the following internal number configurations: (1,1,3) and (1,2,2).

This will not distinguish what the actual card is: only it's face value.
Thanks.

I ended up thinking of it as the number of pairs and the number of non-pairs.

Let's say we have AA and BB and we want to choose 2. We have 3 possible hands: AA, AB, BB.

So Sigma from (r=0 to r=1) (2 choose 1-r)(1+r choose 2r).

(2 choose 1)(1 choose 0) + (2 choose 0)(2 choose 2) = 2+1 = 3