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# Chapter 10 - PowerPoint PPT Presentation

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

Counting Techniques

PermutationsSection 10.2

• A permutation is an arrangement

• of nobjects in a specific order.

• Factorial Formulas

• For any counting n

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

— — — — — —

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.

• The number of permutations of n

• objects taking r objects at a time

• (order is important and nr).

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

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

• 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