1. Destiny is not a matter of chance but a choice
Finding ways to live life to the fullest

2. PERMUTATION
is an arrangement of objects or events in which the order is important. It is also an ordered combination. We can apply the concept of the Fundamental Counting Principle in solving for permutation.

3. Factorial (!)
is commonly used in permutation. For every whole number n,
n >0, the product n(n – 1)(n – 2) ●, … , ●(2)(1) is called n factorial. Zero Factorial (0!) is defined to be one.
4! = (4)(3)(2)(1) = 24
7! = (7)(6)(5)(4)(3)(2)(1) = 5040
2! = (2)(1) = 2
0! = 1

4. nPn = n(n – 1)(n – 2)(n – 3)● , … , ● (3)(2)(1) = n!
The number of permutations of n things taken n at a time.
Example:
5P5 = 5! = (5)(4)(3)(2)(1) = 120
In how many ways can 8 colored pencils be arranged in a box?
Solution:
8P8 = 8! = (8)(7)(6)(5)(4)(3)(2)(1)
= ways

5. nPr = n(n – 1)(n – 2)(n – 3) ● , … , ● (n – r + 1) = 𝒏! 𝒏−𝒓 !
= 𝒏! 𝒏−𝒓 !
The number of permutations on n things taken r at a time without repetition of the objects from the set of n objects.
Example:
In how many ways can a president and a vice president be chosen from a club with 12 members?
Solution: Note that there are 2 positions to be filled in and 12 members can be chosen to fill up the position, hence:

6. nPr = n(n – 1)(n – 2)(n – 3) ● , … , ● (n – r + 1) = 𝒏! 𝒏−𝒓 !
= 𝒏! 𝒏−𝒓 !
12P2 = 𝟏𝟐! 𝟏𝟐−𝟐 ! = 𝟏𝟐! 𝟏𝟎 !
= 𝟏𝟐 𝟏𝟏 𝟏𝟎 (𝟗)(𝟖)(𝟕)(𝟔)(𝟓)(𝟒)(𝟑)(𝟐)(𝟏) (𝟏𝟎)(𝟗)(𝟖)(𝟕)(𝟔)(𝟓)(𝟒)(𝟑)(𝟐)(𝟏)
= (12)(11) = 132 ways or
(12)(11) where in 12 members can be chosen for president and 11 will be left for vice president since any one member of the 12 members has been chosen already as president. We have 132 ways.

7. Note:
In your calculator, you can find the value of 12P2 by pressing the number 12 and then the “P” then the number 2.

8. P = (n – 1)! = (n – 1)(n – 2) ●, … , ● (2)(1)
The number of permutations on n different things in a circular formation.
P = (n – 1)! = (n – 1)(n – 2) ●, … , ● (2)(1)
Example:
How many ways can 10 people be seated in a circular table?
Solution: P = (10 -1)! = 9!
= (9)(8)(7)(6)(5)(4)(3)(2)(1)
= 362,880 ways

9. P = 𝒏! 𝒂!𝒃!𝒄!…
The number of permutations of a thing where a things are alike, b things are alike and c things are alike, and so forth…
Example:
How many permutations can be made for the word MISSISSIPPI?
Solution: Note that there are 11 letters in the word MISSISSIPPI, 1 M, 4 I’s, 4 S’ , 2 P’s, hence:

10. P = 𝒏! 𝒂!𝒃!𝒄!…
Note that there are 11 letters in the word MISSISSIPPI, 1 M, 4 I’s, 4 S’ , 2 P’s, hence:
P = 𝟏𝟏! 𝟒!𝟒!𝟐!𝟏!
= 𝟏𝟏 𝟏𝟎 𝟗 𝟖 𝟕 𝟔 𝟓 (𝟒)(𝟑)(𝟐)(𝟏) 𝟒 𝟑 𝟐 𝟏 𝟒 𝟑 (𝟐)(𝟏)(𝟐)(𝟏)(𝟏)
= (11)(10)(9)(7)(5)= 34,650 ways

11. Questions?

12. The SBC Club organizes a car race in which four cars A,B,C, and D are entered:
A) how many ways can the race be finished if there are no ties?
B) How many ways can the first two positions come in if there are no ties?

13. C) 11P3
D) (8P2)(3P2)

14. Quiz 07/03/2012 BOX your final answer.
1) Find the number of different arrangements of the set of five letters
L, I , V, E, S,
i) taken two at a time
ii) taken three at a time
iii) taken four at a time
2) How many permutations are there in the set (A, B, C, D, E, F)
3) In how many ways can 6 people be arranged in a round table?

15. 4) In how many ways can the following symbols be arranged: ф, Ω, ℓ, ∆, ∂,∏, ¥?
5) Find the number of permutations in the word PENNSYLVANIA.