Number of possibilities to draw from a card deck isn't an integer - where's my error? I have a deck of 40 cards containing A,K,Q,J,10 of the 4 suits, each twice. I want to calculate the number of possibilities to draw 4 cards ignoring the suit.
For the first card, I have 5 possibilities, for the second also, and so on. So I get 5^4 possibilities.
Ignoring the suit means dividing out the permutations of a given draw. For the first card, I have 4 possibilities, for the second 3, and so on. So I need to divide by 4!
But 5^4/4! = 625/24 = 26.04 isn't an integer! Where's my error?
Thanks!
 A: Basically you’ve incorrectly identified the nature of the problem.
Let $n_A,n_K,n_Q,n_J$, and $n_{10}$ be the numbers of aces, kings, queens, jacks, and tens in your set of $4$ cards. In effect you’re asking for the number of solutions in non-negative integers to the equation
$$n_A+n_K+n_Q+n_J+n_{10}=4\;.$$
This is a standard stars and bars problem, and the solution is given by the binomial coefficient
$$\binom{4+5-1}{5-1}=\binom84=70\;.$$
A fairly clear explanation of the formula and its derivation is given in the linked article.
A: first of all, 5 cards in 4 suits is 20, not 40. So either you have all cards double, or you have a typo.
Second, 5^4 is only if you put back the cards, otherwise you'd have 20*19*18*17, and you'd have to find a different way of ignoring suits.
A: *

*Four of the same rank: number of possible ranks so $5$

*Three of one rank, one of a different rank: so $5\times 4=20$ 

*Two of one rank, two of another: so ${5 \choose 2} = 10$

*Two of one rank, one of another, one of a third: so $5 \times {4 \choose 2} =30$

*All four different ranks: ${5\choose 4} = 5$


$5+20+10+30+5=70$ possible patterns but they are not equally probable  
