4-2 Factorials and Permutations Imagine 3 animals running a race: How many different finish orders could there be? D H S FINISHFINISH.

4-2 Factorials and

Permutations

Imagine 3 animals running a race:

How many different finish orders could there be?

D

H

S

FINISH

1st

D

2nd 3rd Order

H

S

H

S

D

S

H

D

S

H

S

D

D

H

DHS

DSH

HDS

HSD

SHD

SDH

THEREFORE:

There are 6 possible permutations (ordered lists) for the race.

This technique will be too cumbersome for questions with any complexity……

So….

Another way:

1st 2nd 3rd

How many choices are there for first place? 3

3

Second place? 2

2

Third place? 1

1

3 X 2 X 1 = 6

(This can be compressed even further)

Note:

3 X 2 X 1 can be compressed into Factorial Notation: 3!

n! = n X (n – 1) X (n – 2) X … X 3 X 2 X 1

Ex: 5! = 5 X 4 X 3 X 2 X 1 = 120

Simplify (on board)

8! =

10! =7!

The senior choir has a concert coming up where they will

perform 5 songs. In how many different orders can they sing

the songs?

In how many ways could 10 questions on a test be arranged if

a) there are no limitations

b) the Easiest question and the most Difficult question are side by side

c) E and D are never side by side

a) No limitations

X X X X X X X X X10 9 8 7 6 5 4 3 2 1 =10!

b) E and D are side by side

X X X X X X X X XE D = 9! X 2

c) E and D are never side by side

10! – 9! X 2

Permutation (when order matters)

A permutation is an ordered arrangement of objects (r) selected from a set (n).

P(n,r) (also written as nPr) represents the number of permutations possible in which r objects from a set of n different objects are arranged.

With the 3 animal race, it would have been 3 objects (n = 3), permute 3 objects (r = 3)

P(3,3) or 3P3

How many first, second, and third place finishers can there

be with 5 animals?

5

1st 2nd 3rd

4 3

5 X 4 X 3 = 60 (way too many to tree)

P(5,3) or 5P3

We want to use the factorial notation….

5 animals, 3 spots…

5 X 4 X 3 X 2 X 1

2 X 1

1

1

1

1

= 5 X 4 X 3

= 60

5!2!

= 5!

(5 – 3)!

= n!

(n – r)!

P(n,r)

How many different sequences of 13 cards can be drawn from a

deck of 52?

52P13 = )!39(!3940414243444546474849505152

211095.3

52P13 = )!1352(

!3940414243444546474849505152

Pg 239

[1-4] odd

7,9,10,11

14,15,19,20