If seven cards are drawn in succession without replacement from a standard deck of fifty two cards how many sets of seven cards are possible?

Answer 1

The first card can be any card in the deck, the second can be any one of the remaining cards, and so on.

52 * 51 * 50 * 49 * 48 = 311,875200

Since we do not care about the order of the cards, this is reduced. 5 cards can be arranged in (5!) 120 different ways.

311,875200 / 120 = 2,598960 possible hands. ■

Answer 2

You would have to draw 17 cards from a deck in order to be sure of getting at least five cards in one suit. (Any suite. Many more if you wanted a particular suite.)

There are four suits. You could draw 16 cards and still not have a five-card suit if you got four cards in each suit. However, the seventeenth card would have to turn one suit into a five-card suit.

Answer 3

52*51*50*49*48 / (5*5*3*2*1)

Answer 4

20358520.00

Answer 5

133,784,560

Answer 6

Solution

Use the Combination Rule.

When the order in which the objects are chosen is not important, the number of ways of choosing r objects from n distinct objects is n C r = n!/r!(n-r)!

There are 6 cards or objects being chosen from a group of 52 cards or objects. The drawing is done without replacement, and order is not important with respect to a hand of cards.

n C r = n!/r!(n-r)! = 52 C 6 = 52!/6!(52-6)! = (52 x 51 x 50 x 49 x 48 x 47 x 46)/(6 x 5 x 4 x 3 x 2 x 1 (46)!)

The 46! cancels out in the top and bottom so the remaining multiplication leaves you with 52C6 = 14658134400/720 , which divides to 20358520

Answer 7

20000