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Chapter 10

Chapter 10. Counting Techniques. Permutations Section 10.2. Permutations. A permutation is an arrangement of n objects in a specific order. Factorial Notation. Factorial Formulas For any counting n. Exercises:. In how many ways can 4 people be seated in a row?.

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Chapter 10

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  1. Chapter 10 Counting Techniques

  2. PermutationsSection 10.2

  3. Permutations • A permutation is an arrangement • of nobjects in a specific order.

  4. Factorial Notation • Factorial Formulas • For any counting n

  5. 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21 = 6! = 720 — — — — — —

  6. Exercises: In how many ways can 4 people be seated in a circle? Formula: (n – 1)! (4 – 1)! = 3! = 321 = 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: 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

  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) = 25 – n(no boys(all girls)) = 25 – 1 = 31 • Out of 5 children, in how many ways • can a family have at least2 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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