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: 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
— — — — — —

6. Exercises: Formula: (n – 1)! 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. 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

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 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