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Chapter 10 Counting Techniques. Permutations Section 10.2.

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Presentation on theme: "Chapter 10 Counting Techniques. Permutations Section 10.2."— Presentation transcript:

1 Chapter 10 Counting Techniques

2 Permutations Section 10.2

3 Permutations A permutation is an arrangement of n objects in a specific order.

4 Factorial Notation Factorial Formulas For any counting n

5 Exercises: = 4! = = 6! = 720 In how many ways can 4 people be seated in a row? 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 Exercises: In how many ways can 4 people be seated in a circle? Formula: (n – 1)! (4 – 1)! = 3! = = 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.

7 Permutation Formula The number of permutations of n objects taking r objects at a time (order is important and n r).

8 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

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

10 Complementary Counting Principle

11 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) = – n(no boys(all girls)) = – 1 = 31 END 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) = – [n(no boys) + n(1 boy)] = – [1 + 5] = 26


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