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EXAMPLE 1

CARDS. A standard deck of 52 playing cards has 4 suits with 13 different cards in each suit. If the order in which the cards are dealt is not important, how many different 5- card hands are possible?. In how many 5 -card hands are all 5 cards of the same color?. EXAMPLE 1.

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EXAMPLE 1

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  1. CARDS A standard deck of 52 playing cards has 4 suits with 13 different cards in each suit. If the order in which the cards are dealt is not important, how many different 5-card hands are possible? In how many 5-card hands are all 5 cards of the same color? EXAMPLE 1 Find combinations

  2. The number of ways to choose 5cards from a deck of 52cards is: 52! 52 51 50 49 48 47! 52C5 = = 47! 5! 47! 5! EXAMPLE 1 Find combinations SOLUTION = 2,598,960

  3. 2! 26! 2C1 26C5 = 1! 1! 21! 5! 26 25 24 23 22 21! 2 = 1 1 21! 5! For all 5 cards to be the same color, you need to choose 1of the 2colors and then 5of the 26cards in that color. So, the number of possible hands is: EXAMPLE 1 Find combinations = 131,560

  4. How many different sets of exactly 2 comedies and 1 tragedy can you read? How many different sets of at most 3 plays can you read? EXAMPLE 2 Decide to multiply or add combinations THEATER William Shakespeare wrote 38 plays that can be divided into three genres. Of the 38 plays, 18 are comedies, 10 are histories, and 10 are tragedies.

  5. You can choose 2of the 18comedies and 1of the 10tragedies. So, the number of possible sets of plays is: 18! 10! 18C2 10C1 = 16! 2! 9! 1! 10 9! 18 17 16! = 9! 1 16! 2 1 = 153 10 EXAMPLE 2 Decide to multiply or add combinations SOLUTION = 1530

  6. You can read 0, 1, 2, or 3plays. Because there are 38plays that can be chosen, the number of possible sets of plays is: 38C0+ 38C1+ 38C2+38C3 = 1 + 38 + 703 + 8436 EXAMPLE 2 Decide to multiply or add combinations = 9178

  7. EXAMPLE 3 Solve a multi-step problem BASKETBALL During the school year, the girl’s basketball team is scheduled to play 12 home games. You want to attend at least 3 of the games. How many different combinations of games can you attend? SOLUTION Of the 12home games, you want to attend 3games, or 4games, or 5games, and so on. So, the number of combinations of games you can attend is: 12C3+ 12C4+ 12C5+…+ 12C12

  8. (12C0+ 12C1+ 12C2) 212– = 4096 – (1 + 12 + 66) EXAMPLE 3 Solve a multi-step problem Instead of adding these combinations, use the following reasoning. For each of the 12games, you can choose to attend or not attend the game, so there are 212total combinations. If you attend at least 3 games, you do not attend only a total of0, 1, or 2games. So, the number of ways you can attend at least 3 games is: = 4017

  9. 8C3 1. ANSWER 56 for Examples 1, 2 and 3 GUIDED PRACTICE Find the number of combinations.

  10. 10C6 2. ANSWER 210 for Examples 1, 2 and 3 GUIDED PRACTICE Find the number of combinations.

  11. 7C2 3. ANSWER 21 for Examples 1, 2 and 3 GUIDED PRACTICE Find the number of combinations.

  12. 14C5 4. ANSWER 2002 for Examples 1, 2 and 3 GUIDED PRACTICE Find the number of combinations.

  13. ANSWER 5400 sets WHAT IF?In Example 2, how many different sets of exactly 3 tragedies and 2 histories can you read? for Examples 1, 2 and 3 GUIDED PRACTICE

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