Five cards are chosen from a standard 52 card deck. How many hands include either an Ace or a Spade.

U have 13 spades and 3 extra aces for a total of 16 cards.

U can make 10 -5 card hands with deck. This will give u a possibility of 10 hands with at lest one spade or and ace

Chuck

I'm afraid creahands's answer is incorrect. The way to approach this is as follows:

As already pointed out, there are 16 cards that satisfy the condition of being either an ace or a spade. That means there are 52-16 = 36 cards that do not satisfy either condition. Here are two ways to continue:

1. First determine how many hands can be dealt with exactly one spade/ace card and 4 others. This would be C(5,1)x16x36*35*34*33. Then add the number of hands that can have two spade/ace cards and three others, plus the number of hands with three spade/ace and 2 others, plus hands with 4 space/ace and 1 other, plus hands with 5 spade/ace and 0 others.

2. An easier way: determine how many total 5-card hands there are in a 52-card deck, then subtract the number of 5-card hands that consist only of non spade/ace cards:

P(52,5) - P(36,5)

If the question is how many combination can be dealt, then my answer is incorrect.

If, however, the question is how many can dealt from a 52 card deck then answer is correct. U can only get 10 hands out of a 52 card deck.

Chuck

If the problem meant for the hand to be sat aside, and then pick another hand from what was left, it would say so. And I've never seen a question wanting to know what it is you were stating.

It wants combinations, i.e. one hand could have a crossover with another hand.

Speaking of which, shouldn't that be combinations and not permutations?

Isn't the question asking for only one spade or one ace?

So, this comes simply to:

^5 C_4 (16)(36\times35\times34\times33)

I wondered about that too. A little on the interpretive side. It says "an" ace or "a" spade. As though they may mean just one of them.