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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

Chapter 10

Counting Techniques


Chapter 10

PermutationsSection 10.2


Chapter 10

Permutations

  • A permutation is an arrangement

  • of nobjects in a specific order.


Chapter 10

Factorial Notation

  • Factorial Formulas

  • For any counting n


Chapter 10

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

— — — — — —


Chapter 10

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.


Chapter 10

Permutation Formula

  • The number of permutations of n

  • objects taking r objects at a time

  • (order is important and nr).


Chapter 10

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


Chapter 10

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



Chapter 10

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