The document defines permutation as an arrangement of objects in a specific order. It provides examples of permutations with different numbers of objects. It discusses factorial notation for calculating permutations. It also covers circular permutations and permutations with repetition. Key concepts covered include: - Permutation formulas for calculating arrangements of objects - Examples of permutations with chairs, books, family members, etc. - Factorial notation and its use in permutation calculations - The difference between linear and circular permutations - Permutations with repetition when objects are not all unique

1. Permutation

2. ×Permutation is an arrangement of objects in a specific
order.
× Permutation also refers to any one of all possible
arrangements of the elements of the given set.
× Permutation is when the order or arrangement is
IMPORTANT.

3. Example #1
If there are three students
and three vacant chairs in
front how many possible
arrangements can we
arrange the students?

4. Therefore
there are 6
possible
arrangement.
S1-S2-S3
S3-S1-S2
S2-S1-S3
S1-S3-S2
S2-S3-S1
S3-S2-S1

5. !!!
Instead of listing all the
possible arrangement
there is an easy way to
know how many
possible arrangements
are there.

6. P(n,r)= n(n-1)(n-2)….(n-r+1)

7. Example #2:
How many different
ways can 6 different
books be arranged on
a shelf?

8. Given: n= 6 books
P(n,r)= n(n-1)(n-2)….(n-r+1)
P(n,r)= 6 (6-1)(6-2)(6-3)(6-4)(6-5)
P(n,r)= 6 (5)(4)(3)(2)(1)
P(n,r)= 720 Simply
P(n,r)= 6! = 720

9.  Means to multiply a series of descending natural numbers.
 It's a shorthand way of writing numbers
 the product of all positive integers less than or equal to n
n n!
1 1 1 1
2 2 × 1 = 2 × 1! = 2
3 3 × 2 × 1 = 3 × 2! = 6
4 4 × 3 × 2 × 1 = 4 × 3! = 24
5 5 × 4 × 3 × 2 × 1 = 5 × 4! = 120

10. Suppose you work at a music
store and have four CDs you
wish to arrange from left to right
on a display shelf. The four CDs
are hip-hop, country, rock, and
alternative (shorthand: H, C, R, A).
How many options do you have?

11. Solution:
Given: n= 4
P(n,r)= n(n-1)(n-2)….(n-r+1)
P(n,r)= 4 (4-1)(4-2)(4-3)
P(n,r)= 4(3)(2)(1)
P(n,r)= 24
Therefore , the CD’s can be
arranged in 24 different ways.

12. A father, mother, 2 boys, and 3 girls are asked
to line up for a photograph. Determine the
number of ways they can line up if:
a. there are no restrictions
b. the parents stand together
c. all the females stand together

13. n= 7!
n= 7 x 6 x 5 x 4 x 3 x 2 x 1
n= 5040 Therefore, the family
members can be line up in
5040 ways.

14. n= 6! 2!
n= (6x 5 x 4 x 3 x 2 x 1) (2 x 1)
n= (720)(2)
n= 1440
Therefore, the family
members can be line up in
1440 ways if the parents
stand together.

15. n= 4! 4!
n= (4 x 3 x 2 x 1)(4 x3 x 2 x 1)
n= (24)(24)
n= 576
Therefore, there are 576
different ways the family
can line up if the females
stand together.

16. Aaron has 3 math books, 4 science books,
and 5 history books. How many different ways
can these books be arranged on a shelf if
books of the same subject must be kept
together?

17. n= (3!) x (5!)(4!)(3!)
n= (6) x (120)(24)(6)
n= (6) x (17, 280)
n= 103 680

18. The Racing Club organizes a race in which 5
cars A, B, C, D, and E are joined. How many ways
can the first two positions be filled if there are
no ties?

19. 20
4
5
2
5 
 x
P
* we used to represent the number of
possible permutations of 5 things when only
2 out of the 5 things are taken.

20. Place your screenshot here
the number of
permutations that
can be made with
n things taken r at
a time
r
n P

21. The number of permutations of n
things taken rat a time is given by:
= n(n-1)(n-2)…(n-r+1)
Look at the last factor:
n-2 = n- (3-1)
= n- (r-1)
= n-r+1
NOTE: n ≥ r
=n(n-1)(n-2)…(n-r+1)
)!
(
!
r
n
n
Pr
n



22. In a school club, there are 5 possible choices
for the president, a secretary, a treasurer, and
an auditor. Assuming that each of them is
qualified for any of these position. In how many
ways can the 4 officers be selected?

23. r
n P
= n(n-1)(n-2)…(n-r+1)
= 5 (5-1)(5-2)(5-3)
= 5 x 4 x 3 x 2
= 120 ways
=
𝑛!
𝑛−𝑟 !
=
5!
5−4 !
=
5 𝑥 4 𝑥 3 𝑥 2 𝑥 1
1 𝑥 1
=
120
1
= 120

24. The Open Minded Band has 20 songs to
perform in a concert. At the upcoming
Battle of the Bands,they will play 2 songs.
In how many different orders can they
perform two of their songs?

25. 20P2 =
20!
20−2 !
20P2 = 380

26. The number of permutations of n things
taken n at a time is given by:
= n(n-1)(n-2)…(3)(2)(1)= n!
Pn
n

27. The Racing Club organizes a race in which
5 cars A, B, C, D, and E are joined. How
many possible race results if there are no
ties?

28. 120
1
2
3
4
5
5
5 
 x
x
x
x
P

29. Circular
Permutation

30. Place your screenshot here
 Are the permutation of
objects when they are
arranged in a crcular
pattern
Is a special case of
permutation where the
arrangements of things
is in a circular pattern.

31. Place your screenshot
here
Example # 10:
In how many ways can 3
people be seated around a
circular table?

32. Solution:
Given: n= 3
Therefore, 3!
3!= 3 X 2 X 1
3!= 6
A-B-C
A-C-B
B-C-A
B-A-C
C-A-B
C-B-A

33. Observe that all the arrangements falling
on the same column are just the same
because the 3 people are supposed to be
seated around a circular table. There are 6
arrangements.

34. A
C
B
A
C B
there are only 2 possible
permutation in arranging
3 persons at a round
table.

35. Therefore the circular
permutations, P of 3 objects
is:
n
n
P
!

P=
3!
3
P=
3 𝑋 2 𝑋 1
3
P=
6
3
P= 2

36. Did you know that,
n
n
P
!

P= (n – 1)!

37. In how many different ways can you
arranged 8 figures on a circular
shelf?

38. Given: n= 8
P= (n-1)!
P= (8-1)!
P= 7!
P= 5, 040
P=
𝑛!
𝑛
P=
8!
8
P=
40 320
8
P=5 040

39. Mr. Kesler is at a picnic that has a circular
revolving condiment server. Six
condiments are placed on the table. How
many ways can the condiments be
arranged?

40. Given: n= 6
P= (n-1)!
P= (6-1)!
P= 5!
P= 120
P=
𝑛!
𝑛
P=
6!
6
P=
720
6
P=120

41. Eleven people are to be seated at a round table
where one person is seated closest to the exit.
How many possible arrangements of people
relative to the exit are possible?

42. Given: n= 11
P= (n-1)!
P= (11-1)!
P= 10!
P= 3, 628, 800
P=
𝑛!
𝑛
P=
11!
11
P=
39,916,800
11
P=3, 628, 800

43. Permutation
with
REPETITION

44. Example # 14:
How many distinguishable permutation are
there for the letters of the word TENNESSEE
taken all together?

45. Solution:
Given:
Letters in all= 9
E’s= 4
N’s= 2
S’s= 2
P=
𝑛!
𝑛1!𝑛2!..
P=
9!
4!2!2!
P=
362,880
96
P= 3780

46. Example # 15:
Find the number of distinguishable
permutations of the letters in the word
“ BASKETBALL”.

47. Solution:
!
2
!
2
!
2
!
10

P
Given:
All letters: 10
B’s= 2
A’s = 2
L’s= 2
P=
3, 628, 800
8
P= 453, 600 ways

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