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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