PERMUTATION OF AN OBJECT

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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