Simple Combinatorics - drawing aces from a deck of cards Suppose you have a deck of cards which contains 52 cards with 4 different sets which consist of 13 cards each.
We write a series of $n$ cards like this: 
We draw a card, write it as the next cards in our series of cards, put the cards back to the deck and shuffle it. we repeat on this process $n$ times, in how many ways every "Ace" cards will be picked at least once?
Thanks all!
 A: There are probably better "prettier" ways to evaluate this, but inclusion-exclusion over the various aces is not too hard either.
The method here is to evaluate how many ways we can break the constraint (of having every Ace picked at least once), then subtract that from the total possibilities of $52^n$.
The number of ways in which any one given ace can be avoided is $51^n$. The number of ways in which any two given aces can be avoided is $50^n$, the number of ways for any three avoided aces to be avoided is $49^n$, and amazingly the number of ways for all four aces to be avoided is $48^n$.
Of course when we consider options that miss one ace, some of those possibilities also miss some of the other aces. so we have to account for double-counting and double corrections etc. with inclusion-exclusion.
Using inclusion-exclusion, the possibilities for not getting all four aces then is $$ {4 \choose 1}51^n - {4 \choose 2}50^n + {4 \choose 3}49^n - {4 \choose 4}48^n $$
and the possibilities for getting at least one ace of each suit are$$ 52^n- {4 \choose 1}51^n + {4 \choose 2}50^n - {4 \choose 3}49^n + {4 \choose 4}48^n $$
Testing with n = 1,2,3,4 we happily get 0,0,0,24.
