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1
Chapter 10 Counting Techniques

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Permutations Section 10.2

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**Permutations A permutation is an arrangement**

of n objects in a specific order.

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Factorial Notation Factorial Formulas For any counting n

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**Exercises: In how many ways can 4 people be seated in a row?**

4 3 2 1 = 4! = 24 If 6 horses are in a race and they all finish with no ties, in how many ways can the horses finish the race? 6 5 4 3 1 = 6! = 720 — — — — — —

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**Exercises: Formula: (n – 1)! In how many ways can 4 people**

be seated in a circle? Formula: (n – 1)! (4 – 1)! = 3! = 321 = 6 Notice: The answer is not the same as standing in a row. The reason is everyone could shift one seat to the right (left) but they would still be sitting in the same order or position relative to each other.

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**Permutation Formula The number of permutations of n**

objects taking r objects at a time (order is important and n r).

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**Exercise: A basketball coach must choose 4 players to**

play in a particular game. (The team already has a center.) In how many ways can the 4 positions be filled if the coach has 10 players who can play any position? 10 nPr 4 = 10 x 9 x 8 x 7 = 5040

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**Assume the cards are drawn**

Exercises: Assume the cards are drawn without replacement. In how many ways can 3 hearts be drawn from a standard deck of 52 cards? 13 nPr 3 = 13 x 12 x 11 = 1716 In how many ways can 2 kings be drawn from a standard deck of 52 cards? 4 nPr 2 = 4 x 3 = 12

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**Complementary Counting Principle**

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**Exercises: Out of 5 children, in how many ways**

can a family have at least 1 boy? n(A) = n(U) – n(A' ) n(at least 1 boy) = 25 – n(no boys(all girls)) = 25 – 1 = 31 Out of 5 children, in how many ways can a family have at least 2 boys? n(A) = n(U) – n(A' ) n(at least 2 boys) = 25 – [n(no boys) + n(1 boy)] = 25 – [1 + 5] = 26 END

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Chapter 8 Counting Principles: Further Probability Topics Section 8.2 Combinations.

Chapter 8 Counting Principles: Further Probability Topics Section 8.2 Combinations.

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