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6.1 Use Combinations and Binomial Theorem 378 What is a combination? How is it different from a permutation? What is the formula for n C r ?

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In the last section we learned counting problems where order was important For other counting problems where order is NOT important like cards, (the order youre dealt is not important, after you get them, reordering them doesnt change your hand) For other counting problems where order is NOT important like cards, (the order youre dealt is not important, after you get them, reordering them doesnt change your hand) These unordered groupings are called Combinations These unordered groupings are called Combinations

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A Combination is a selection ofr objects from a group of n objects where order is not important

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Combination of n objects taken r at a time The number of combinations of r objects taken from a group of n distinct objects is denoted by n C r and is: The number of combinations of r objects taken from a group of n distinct objects is denoted by n C r and is:

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For instance, the number of combinations of 2 objects taken from a group of 5 objects is For instance, the number of combinations of 2 objects taken from a group of 5 objects is 2

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Finding Combinations In a standard deck of 52 cards there are 4 suits with 13 of each suit. In a standard deck of 52 cards there are 4 suits with 13 of each suit. If the order isnt important how many different 5-card hands are possible? If the order isnt important how many different 5-card hands are possible? The number of ways to draw 5 cards from 52 is The number of ways to draw 5 cards from 52 is = 2,598,960

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In how many of these hands are all 5 cards the same suit? You need to choose 1 of the 4 suits and then 5 of the 13 cards in the suit. You need to choose 1 of the 4 suits and then 5 of the 13 cards in the suit. The number of possible hands are: The number of possible hands are:

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How many 7 card hands are possible? How many of these hands have all 7 cards the same suit? How many of these hands have all 7 cards the same suit?

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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. a. How many different sets of exactly 2 comedies and 1 tragedy can you read? a. 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 118 C 2 16! 2! 18! = 18 17 16! 16! 2 1 9! 1 10 9! = 1530

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b. How many different sets of at most 3 plays can you read? b. 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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When finding the number of ways both an event A and an event B can occur, you multiply. When finding the number of ways both an event A and an event B can occur, you multiply. When finding the number of ways that an event A OR B can occur, you +. When finding the number of ways that an event A OR B can occur, you +.

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Deciding to + or * A restaurant serves omelets. They offer 6 vegetarian ingredients and 4 meat ingredients. A restaurant serves omelets. They offer 6 vegetarian ingredients and 4 meat ingredients. You want exactly 2 veg. ingredients and 1 meat. How many kinds of omelets can you order? You want exactly 2 veg. ingredients and 1 meat. How many kinds of omelets can you order?

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Suppose you can afford at most 3 ingredients How many different types can you order? How many different types can you order? You can order an omelet with 0, or 1, or 2, or 3 items and there are 10 items to choose from. You can order an omelet with 0, or 1, or 2, or 3 items and there are 10 items to choose from.

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Counting problems that involve at least or at most sometimes are easier to solve by subtracting possibilities you dont want from the total number of possibilities. Counting problems that involve at least or at most sometimes are easier to solve by subtracting possibilities you dont want from the total number of possibilities.

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Subtracting instead of adding: A theatre is having 12 plays. You want to attend at least 3. How many combinations of plays can you attend? A theatre is having 12 plays. You want to attend at least 3. How many combinations of plays can you attend? You want to attend 3 or 4 or 5 or … or 12. You want to attend 3 or 4 or 5 or … or 12. From this section you would solve the problem using: From this section you would solve the problem using: Or…… Or……

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For each play you can attend you can go or not go. For each play you can attend you can go or not go. So, like section 12.1 it would be 2*2*2*2*2*2*2*2*2*2*2*2 =2 12 So, like section 12.1 it would be 2*2*2*2*2*2*2*2*2*2*2*2 =2 12 And you will not attend 0, or 1, or 2. And you will not attend 0, or 1, or 2. So: So:

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Basketball During the school year, the girls 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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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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6.1 Assignment Day 1 Page 382, 3-18 all, 38-40 all Page 382, 3-18 all, 38-40 all

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6.1 Use Combinations and Binomial Theorem Day 2 What is Pascals Triangle? What is the Binomial Theorem? How is Pascals Triangle connected to the binomial theorem?

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Pascals Triangle If you arrange the values of n C r in a triangular pattern in which each row corresponds to a value N, you get what is called Pascals triangle. Pascals triangle is named after the French mathematician Blaise Pascal (1623-1662) If you arrange the values of n C r in a triangular pattern in which each row corresponds to a value N, you get what is called Pascals triangle. Pascals triangle is named after the French mathematician Blaise Pascal (1623-1662)

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The Binomial Theorem 0 C 0 0 C 0 1 C 0 1 C 1 1 C 0 1 C 1 2 C 0 2 C 1 2 C 2 2 C 0 2 C 1 2 C 2 3 C 0 3 C 1 3 C 2 3 C 3 3 C 0 3 C 1 3 C 2 3 C 3 4 C 0 4 C 1 4 C 2 4 C 3 4 C 4 4 C 0 4 C 1 4 C 2 4 C 3 4 C 4 Etc… Etc…

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Pascal's Triangle! 1 1 1 1 1 1 2 1 1 2 1 1 3 3 1 1 3 3 1 1 4 6 4 1 1 4 6 4 1 1 5 10 10 5 1 1 5 10 10 5 1 Etc… Etc… This describes the coefficients in the expansion of the binomial (a+b) n This describes the coefficients in the expansion of the binomial (a+b) n

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(a+b) 2 = a 2 + 2ab + b 2 (1 2 1) (a+b) 2 = a 2 + 2ab + b 2 (1 2 1) (a+b) 3 = a 3 (b 0 )+3a 2 b 1 +3a 1 b 2 +b 3 (a 0 ) (1 3 3 1) (a+b) 3 = a 3 (b 0 )+3a 2 b 1 +3a 1 b 2 +b 3 (a 0 ) (1 3 3 1) (a+b) 4 = a 4 +4a 3 b+6a 2 b 2 +4ab 3 +b 4 (1 4 6 4 1) (a+b) 4 = a 4 +4a 3 b+6a 2 b 2 +4ab 3 +b 4 (1 4 6 4 1) In general… In general…

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(a+b) n (n is a positive integer)= n C 0 a n b 0 + n C 1 a n-1 b 1 + n C 2 a n-2 b 2 + …+ n C n a 0 b n n C 0 a n b 0 + n C 1 a n-1 b 1 + n C 2 a n-2 b 2 + …+ n C n a 0 b n =

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(a+3) 5 = 5 C 0 a 5 3 0 + 5 C 1 a 4 3 1 + 5 C 2 a 3 3 2 + 5 C 3 a 2 3 3 + 5 C 4 a 1 3 4 + 5 C 5 a 0 3 5 = 5 C 0 a 5 3 0 + 5 C 1 a 4 3 1 + 5 C 2 a 3 3 2 + 5 C 3 a 2 3 3 + 5 C 4 a 1 3 4 + 5 C 5 a 0 3 5 = 1a 5 + 15a 4 + 90a 3 + 270a 2 + 405a + 243 1a 5 + 15a 4 + 90a 3 + 270a 2 + 405a + 243

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(a+3) 5 = 1a530+5a431+10a332+10a233+5a134 +1a035 1a5 + 15a4 + 90a3 + 270a2 + 405a + 243 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1

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School Clubs The 6 members of a Model UN club must choose 2 representatives to attend a state convention. Use Pascals 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 Pascals triangle by adding numbers from the previous row.

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n = 5 (5th row ) 1 5 10 10 5 1 n = 6 (6th row ) 1 6 15 20 15 6 1 6C06C0 6C16C16C26C26C36C36C46C46C56C56C66C6 ANSWER The value of 6 C 2 is the third number in the 6 th row of Pascals triangle, as shown above. Therefore, 6 C 2 = 15. There are 15 combinations of representatives for the convention.

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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 + 3x 4 y + 3x 2 y 2 + y 3

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

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Find the coefficient of x 5 in the expansion of (x – 3) 7. From the binomial theorem, you know the following: Each term in the expansion has the form 7 C r (x) 7 – r (–3) r. The term containing x 5 occurs when r = 2 : ANSWERThe coefficient of x 5 is 189. SOLUTION (x – 3) 7 7 C 0 (x) 7 (–3) 0 + 7 C 1 (x) 6 (–3) 1 +... + 7 C 7 (x) 0 (–3) 7 = 7 C r (x) 7 – r (–3) r = (21)(x 5 )(9)= 189x 5

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From the binomial theorem, you know the following: Each term in the expansion has the form 8 C r (2x) 8 – r (5) r. The term containing x 3 occurs when r = 5 : ANSWERThe coefficient of x 3 is 1,400,000. SOLUTION Find the coefficient of x 3 in the expansion of (2x + 5) 8. = (2x + 5) 8 8 C 0 (2x) 8 (5) 0 + 8 C 1 (2x) 7 (5) 1 +... + 8 C 8 (2x) 0 (5) 8 8 C 5 (2x) 3 (5) 5 = (56)(8x 3 )(3125)= 1,400,000x 3

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What is Pascals Triangle? What is Pascals Triangle? An arrangement of the values of n C r in a triangular pattern. What is the Binomial Theorem? What is the Binomial Theorem? A formula that gives the coefficients in the raising of (a+b) to a power. How is Pascals Triangle connected to the binomial theorem? How is Pascals Triangle connected to the binomial theorem? Pascals Triangle gives the same coefficients as the Binomial Theorem.

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Assignment 12.2 Page 382 Page 382 20-30even, 32-34 all,48-50 all

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12.2 Combinations and Binomial Theorem p. 708. In the last section we learned counting problems where order was important For other counting problems.

12.2 Combinations and Binomial Theorem p. 708. In the last section we learned counting problems where order was important For other counting problems.

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