SOLUTION: In the game of Texas hold'em, a player is dealt two cards (called hole cards) from a shuffled deck of 52 playing cards. The order in which the cards are dealt does not matter. 1

Question 1025774: In the game of Texas hold'em, a player is dealt two cards (called hole cards) from a shuffled deck of 52 playing cards. The order in which the cards are dealt does not matter.
1. Find the number of hole cards combination that is possible?
2. How many combinations are there in which both cards are aces?
Can someone help me step by step how to solve this?
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
You may want to review the binomial coefficient operator ("n choose k").

Note that the order of which the two hole cards is dealt is irrelevant in Texas Hold'em (although the ordering of any future cards is!), so use combinations.

1. Choose 2 cards from 52 to represent the hole cards. 52C2 = 52*51/2 = 1326
2. There are four aces, choose 2 cards from 4 aces. 4C2 = 4*3/2 = 6
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