In the card game bridge the game begins by dividing up a deck of 52 cards among four players with each receiving 13. (We can assume that the players are numbered from 1 to 4.) Find the number of ways the following outcomes can happen. (a) Player 1 receives all four aces. (b) Each player receives 13 cards of the same suit. (c) Player 1 and Player 2 together receive all the O cards

Explanation:

Explanation:

Explanation:

In a the card game there are 4 groups N,E,W,S. Each group consist of 13 cards each

N: A,K,Q,J,10,9,8,7,6,5,4,3,2

E: A,K,Q,J,10,9,8,7,6,5,4,3,2

W: A,K,Q,J,10,9,8,7,6,5,4,3,2

S: A,K,Q,J,10,9,8,7,6,5,4,3,2

a) The number of ways player 1 receives 4 aces

Since 52 cards can be arranged in 52! ways

An A's card can be arranged in 4! ways

Number of possible arrangement of A's cards among the 4 player =( 52! / 4! )

So for 1 player to receive all the Ace's there will be (13!/4!) ways

13!/4! = 259459200 ways

b) each player receive 13cards of the same suit

There are 4! different ways to deal the cards so that each player gets all the cards of one suit.

The distribute 4 objects (suits) to the 4 players.

The number of ways to deal out the cards is (52!/13!) X (39!/13!) X (26! / 13!) X (13!/13!). You choose 13 cards to give to the first player, 13 to the second, 13 to the third and 13 to the fourth.

Using the rule for probability, you get an answer of

4! / ((52!/13!) X (39!/13!) X (26! / 13!) X (13!/13!))

answer; ///(d) subdatasheet;