L10.2 New

Algebra 2:

Algebra 2 Chapter 10 Lesson 2

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EXAMPLE 1 Find combinations 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?

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

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

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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. 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?

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

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

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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 12 home games, you want to attend 3 games, or 4 games, or 5 games, and so on. So, the number of combinations of games you can attend is: 12 C 3 + 12 C 4 + 12 C 5 +…+ 12 C 12

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EXAMPLE 3 Solve a multi-step problem Instead of adding these combinations, use the following reasoning. For each of the 12 games, you can choose to attend or not attend the game, so there are 2 12 total combinations. If you attend at least 3 games, you do not attend only a total of 0 , 1 , or 2 games. So, the number of ways you can attend at least 3 games is: 2 12 – ( 12 C 0 + 12 C 1 + 12 C 2 ) = 4096 – (1 + 12 + 66) = 4017

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

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GUIDED PRACTICE for Examples 1, 2 and 3 Find the number of combinations. 2. 10 C 6 210 ANSWER

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

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GUIDED PRACTICE for Examples 1, 2 and 3 Find the number of combinations. 4. 14 C 5 2002 ANSWER

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

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EXAMPLE 4 Use Pascal’s triangle School Clubs The 6 members of a Model UN club must choose 2 representatives to attend a state convention. Use Pascal’s triangle to find the number of combinations of 2 members that can be chosen as representatives. SOLUTION Because you need to find 6 C 2 , write the 6 th row of Pascal’s triangle by adding numbers from the previous row.

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EXAMPLE 4 Use Pascal’s triangle n = 5 (5 th row ) 1 5 10 10 5 1 n = 6 (6 th row ) 1 6 15 20 15 6 1 6 C 0 6 C 1 6 C 2 6 C 3 6 C 4 6 C 5 6 C 6 ANSWER The value of 6 C 2 is the third number in the 6 th row of Pascal’s triangle, as shown above. Therefore, 6 C 2 = 15 . There are 15 combinations of representatives for the convention.

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GUIDED PRACTICE for Example 4 WHAT IF? In Example 4 , use Pascal’s triangle to find the number of combinations of 2 members that can be chosen if the Model UN club has 7 members. 6. ANSWER 21 combinations

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EXAMPLE 5 Expand a power of a binomial sum Use the binomial theorem to write the binomial expansion. ( x 2 + y ) 3 = 3 C 0 ( x 2 ) 3 y 0 + 3 C 1 ( x 2 ) 2 y 1 + 3 C 2 ( x 2 ) 1 y 2 + 3 C 3 ( x 2 ) 0 y 3 = (1)( x 6 )(1) + (3)( x 4 )( y ) + (3)( x 2 )( y 2 ) + (1)(1)( y 3 ) = x 6 + 3 x 4 y + 3 x 2 y 2 + y 3

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EXAMPLE 6 Expand a power of a binomial difference Use the binomial theorem to write the binomial expansion. ( a – 2 b ) 4 = [ a + (–2 b )] 4 = 4 C 0 a 4 (–2 b ) 0 + 4 C 1 a 3 (–2 b ) 1 + 4 C 2 a 2 (–2 b ) 2 + 4 C 3 a 1 (–2 b ) 3 + 4 C 4 a 0 (–2 b ) 4 = (1)( a 4 )(1) + (4)( a 3 )(–2 b ) + (6)( a 2 )(4 b 2 ) + (4)( a )(–8 b 3 ) + (1)(1)(16 b 4 ) = a 4 – 8 a 3 b + 24 a 2 b 2 – 32 ab 3 + 16 b 4

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GUIDED PRACTICE for Examples 5 and 6 Use the binomial theorem to write the binomial expansion. ( x + 3) 5 x 5 + 15 x 4 + 90 x 3 + 270 x 2 + 405 x + 243 ANSWER

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GUIDED PRACTICE for Examples 5 and 6 Use the binomial theorem to write the binomial expansion. ( a + 2 b ) 4 a 4 + 8 a 3 b + 24 a 2 b 2 + 32 ab 3 + 16 b 4 ANSWER

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GUIDED PRACTICE for Examples 5 and 6 Use the binomial theorem to write the binomial expansion. (2 p – q ) 4 16 p 4 – 32 p 3 q + 24 p 2 q 2 – 8 pq 3 + q 4 ANSWER

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GUIDED PRACTICE for Examples 5 and 6 Use the binomial theorem to write the binomial expansion. (5 – 2 y ) 3 –8 y 3 + 60 y 2 – 150 y + 125 ANSWER

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EXAMPLE 7 Find a coefficient in an expansion Find the coefficient of x 4 in the expansion of (3 x + 2) 10 . SOLUTION From the binomial theorem, you know the following: (3 x + 2) 10 = 10 C 0 (3 x ) 10 (2) 0 + 10 C 1 (3 x ) 9 (2) 1 + . . . + 10 C 10 (3 x ) 0 (2) 10 Each term in the expansion has the form 10 C r (3 x ) 10 – r (2) r . The term containing x 4 occurs when r = 6 : 10 C 6 (3 x ) 4 (2) 6 = (210)(81 x 4 )(64) = 1,088,640 x 4 ANSWER The coefficient of x 4 is 1,088,640 .

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GUIDED PRACTICE for Example 7 Find the coefficient of x 5 in the expansion of ( x – 3) 7 . 11. ANSWER 189

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GUIDED PRACTICE for Example 7 12. ANSWER 1,400,000 Find the coefficient of x 3 in the expansion of (2 x + 5) 8 .