Math PPT

1. PERMUTATION
Prepared by: Judy Anne J. Gawat

2. PLEASE PREPARE YOUR
SCIENTIFIC CALCULATOR AS
IT WILL BE USED IN OUR
DISCUSSION.
NOTE

3. PERMUTATION
•An arrangement of objects in a definite
order or the ordered arrangement of
distinguishable objects without allowing
repetitions among objects.

4. PRELIMINARYTASK
•Using your scientific calculator, find the value of the
following;
1.4!
2.3!5!
3.7!
5!
4. 8!
2!4!
5. 7!
3!4!
*
5!
2!
24
720
42
840
2100

5. What is meant by n! (n factorial)
n!= n (n – 1) (n – 2) ( n – 3)…
4! = (4) (3) (2) (1)
4! = 24
Example:

6. • 3! 5!
3! = (3) (2) (1) = 6
5! = (5) (4) (3) (2) (1) = 120
6 x 120 = 720

7. Match Me!
• How many ways can you match the outfits when
you have three pants and three tops on hand?

8. •How many ways can you match the outfits
when you have three pants and three tops
on hand?
•Can you do this systematically? Can do you
do it in less time and become accurate
about it?

9. Use tree diagram

10. Thus, there are nine resulting
outfits one can make out of
the three different pants and
three different tops.

11. Fundamental Counting Principle
(FCP)
•The manner of solving by
multiplying the number of elements
of two or more events to find the
total number of outcomes for those
events to occur.

12. Types of Permutation
1.Permutation of n objects
2.Distinguishable permutation
3.Permutation of n objects taken r at a
time
4.Circular Permutation.

13. Permutation of n objects
The permutation of n objects is equal
to n factorial)

14. How many arrangements are there?
1. Arranging different 3 portraits on a wall
3! = 6

15. 2. arranging 4 persons in a row for a picture taking
4! = 24
3. Arranging 5 different figurines in a shelf
5! = 120
4. Arranging 6 different potted plats in a row
6! = 720
5. Arranging digits of the number 123456789
9! = 362, 880

16. Distinguishable Permutation
There are repeated ( or identical)
objects in the set.

17. How many arrangement are there?
1. Arranging the digits in the number 09778210229
P =
11!
2!2!2!3 !
= 831 600
2. Drawing one by one and arranging in a row 4
identical blue, 5 identical yellow, and 3 identical red
balls in a bag
P =
12!
3!4!5!
= 27 720

18. 4. Arranging these canned goods
P =
3!
3!2!2!
= 1680
3. Arranging the letters in the word LOLLIPOP
P =
𝟏𝟎!
𝟑!𝟒!
= 𝟐𝟓 𝟐𝟎𝟎

19. Permutation of n objects taken r
Permutation of n taken r at a
time where n ≥ r

20. How many arrangements are there?
1. Choosing 3 posters to hang on a wall from 5
posters you are keeping

21. nPr =
𝑛!
𝑛 −𝑟 !
5P3 =
5!
5 −3 !
5P3 =
5!
2!
= 60
Notation: 5P3 / P(5,3)
Calculator: 5 shift nPr 3 =

22. 2.Taking two –letter word, without repetition of
letters, from the letters of the word LOVED.
Example: LO, OE,VD…
nPr =
𝑛!
𝑛 −𝑟 !
5P2 =
5!
5 −2 !
5P2 =
5!
3!
5P2 = 20
Notation:
5P2 / P (5, 2)
Calculator:
5 shift nPr 2 =

23. 3. Pirena, Amihan, Alena , and Danaya competing
for 1st , 2nd , and 3rd places in Math Quiz bee.
nPr =
𝑛!
𝑛 −𝑟 !
4P3 =
4!
4−3 !
4P3 =
4!
1!
4P3 = 24
Example: Danaya – 1st , Alena – 2nd ,
Amihan – 3rd
Notation:
4P3 / P (4, 3)
Calculator:
4 shift nPr 3 =

24. Circular Permutation
If n objects are arranged in a circle, then there
are
𝑛!
𝑛
𝑜𝑟 𝑛 − 1 !
𝑝𝑒𝑟𝑚𝑢𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑛 𝑜𝑏𝑗𝑒𝑐𝑡𝑠 𝑎𝑟𝑜𝑢𝑛𝑑 𝑡ℎ𝑒 𝑐𝑖𝑟𝑐𝑙𝑒.

25. How many arrangements are there?
1. How many ways can 5 people sit
around a circular table?
P = (5 – 1) !
P = 4 !
P = 24

26. Let’s analyze the situation using the illustration.
Then, P =
5!
5
=
5 𝑥 4!
5
= 1 𝑥 4! = 4! = 24
Therefore, five people can be arranged around
in 24 different ways.

27. 2.Ten boys scouts are to be seated around a
campfire. How many ways can they be
arranged?
•Solution:
P =
10!
10
=
10 𝑥 9!
10
P = 9!
P = 362, 880
Therefore, there are
362, 880 ways to
arrange the ten boy
scouts around a
campfire.

28. 3. Eight people are to be seated at a roundtable.
One of them is to be seated close to the window.
How many arrangements are possible?
•Solution:
P = n! = 8! = 40, 320
Therefore, there are 40, 320 arrangements possible.
Note: if n objects on a circle are arranged relative to a fixed
point, then there are n! permutations even though the
objects are on a circle, the permutations are linear since a
reference point has been establish

29. 4. How many different ways can four keys, no two of
which are the same, be arranged on a key-ring that has
a clasp?
• Solution:
P =
𝑁!
2
=
4!
2
P=
24
2
= 12
• Therefore, there are 12 ways to arrange the four keys, no two of
which are the same, on a key -ring
Note; this is no longer a case of an ordinary circular permutation
since objects are arranged with respect to a fixed point, the clasp
and is reflective. In this case there are
𝒏!
𝟐

30. Activity: find the number of permutations of
the given objects in each of the following
situations.
•1. How many seating arrangements are possible for
seven people at a roundtable?
•2. In how many different ways can six keys, no two of
which are the same, be arranged on a key ring that
has no clasp?

31. 1. Find the number of different arrangements
of the set of the six letters JOYFUL:
a)Taken two at a time
b)Taken three at a time
c)Taken six at a time

32. Solution:
Ex: JO, OL , FU….
a)Taken two at a time
nPr =
𝒏!
𝒏 −𝒓 !
6P2 =
𝟔!
𝟔 −𝟐 !
=
𝟕𝟐𝟎
𝟐𝟒
= 30
Ex: JOY, OLU , FUL….
b)Taken three at a time
nPr =
𝒏!
𝒏 −𝒓 !
6P3 =
𝟔!
𝟔 −𝟑 !
=
𝟕𝟐𝟎
𝟔
= 120

33. Ex: JO, OL , FU….
C)Taken six at a time
nPr =
𝒏!
𝒏 −𝒓 !
6P6 =
𝟔!
𝟔 −𝟔 !
=
𝟕𝟐𝟎
𝟏
= 720

34. 2.There are seven finalists for the Math contest, and
medals will be given to the top three finalists. How
many ways are there for the medal winners to be
selected?
• Solution:
nPr =
𝒏!
𝒏 −𝒓 !
7P3 =
𝟕!
𝟕 −𝟑 !
=
𝟓𝟎𝟒𝟎
𝟐𝟒
= 210

35. 3. In how many ways can each of the following
be drawn from a standard deck of 52 cards?
a)Three diamonds
b)Two picture cards

36. Solution:
a.Three diamonds
52P3 =
𝟓𝟐!
𝟓𝟐 −𝟑 !
=
𝟓𝟐!
𝟒𝟗 !
=
𝟓𝟐 (𝟓𝟏)(𝟓𝟎)(𝟒𝟗!)
𝟒𝟗!
= (52)(51)(50)
= 132600

37. b. Two picture cards
•nPr =
𝒏!
𝒏 −𝒓 !
•52P2 =
𝟓𝟐!
𝟓𝟐 −𝟐 !
• =
𝟓𝟐!
𝟓𝟎 !
• =
(𝟓𝟐)(𝟓𝟏)(𝟓𝟎!)
𝟓𝟎 !
•= (52)(51)
•= 2652

38. 4. In how many ways can eight people be seated at a
round table?
Solution:
P = (n – 1) !
P= (8 – 1)!
P = 7!
P = 5040

39. 5. In how many ways can eight keys be
arranged in a key ring?
Solution:
P = 8!
P = 40320

40. 6. How many different ways be the letters of
each word can be arranged?
a.CIRCUIT
b.FULFILLMENT
c.MATHEMATICS