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12.2 Combinations and Binomial Thm p. 708

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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 of r 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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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 w/ 0, or 1, or 2, or 3 items and there are 10 items to choose from. You can order an omelet w/ 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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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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Assignment

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