How many ways can you select 5 cards from a deck of 52 cards?

How many ways can you select 5 cards from a deck of 52 cards?

2598960 different ways
5! = 2598960 different ways to choose 5 cards from the available 52 cards.

How many ways are there to select a 5 card hand from a regular 52-card deck such that the hand contains at least one card in each suit?

So that’s 13C2 * 13C1 * 13C1 * 13C1 * 4 = 685,464 out of the 52C5 = 2,598,960 possible poker hands. So the probability of it happening is 686,464 / 2,598,960 = . 2637+.

How many 5 card hands are possible if all 5 cards are the same suit?

1287 possible
A hand that is a flush must consist of all five cards being of the same suit. Each of the four suits has 13C5 = 1287 possible five-card hands that are all of the same suit.

How many possible hands of 5 cards are there?

2,598,960
Probability of a Full House First, count the number of five-card hands that can be dealt from a standard deck of 52 cards. We did this in the previous section, and found that there are 2,598,960 distinct poker hands. Next, count the number of ways that five cards can be dealt to produce a full house.

How many ways can you draw 5 cards from a standard deck and get a flush?

There are thirteen ranks, and a flush must contain five of them. (There are no duplicate cards in a standard 52-card deck.) That gives possible flushes per suit (the order in which the cards appear does not matter), times 4 suits, or 5148 distinct flushes.

How many 5 cards consists of only hearts?

5!) = 2,598,960 five card hands. There are 13C5 = 1287 hands that consist entirely of hearts; that’s the difference between 2,598,960 and 2,597,673.

How many ways can we have a poker hand of 5 cards?

Probability of Two Pair First, count the number of five-card hands that can be dealt from a standard deck of 52 cards. We did this previously, and found that there are 2,598,960 distinct poker hands.

How many ways can a 5 card hand contain two pair?

If we order the 5-card hand with the two pairs first, we have 13C2 choices for the two numbers showing on the two pairs. Each pair will have two out of four suits. Thus, we have 4C2·4C2 = 6·6 = 36 ways to choose the suits.