Finding the number of permutation in an object

1. Permutation

2. A small padlocked treasure chest
was found in an abandoned island.
In order to open this chest containing
precious jewels, a 4- letter password
without repetition must be unlocked
using the letters A, B, C, and D. If
you are going to list down all the
possible codes, how many codes will
there be?

3. ×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.

4. Permutation
WITHOUT
restrictions

5. Example #1
Find the number of ways to
arrange the letters ABC.

6. By listing
method there
are 6 ways.
A-B-C
A-C-B
B-A-C
B-C-A
C-A-B
C-B-A

7. !!!
Instead of listing all the
possible arrangement
there is an easy way to
know how many
possible arrangements
are there.

8.  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

9. Solution:
Given: n= 3
n! = n(n-1)(n-2)….(n-r+1)
3! = 3 (3-1)(3-2)
3! = (3)(2)(1)
3! = 6
Therefore , letters ABC can
arrange in 6 ways..

10. Try this!
5!
2!
8!
5x4x3x2x1
2x1
8x7x6x5x4x3x2x1
= 120
= 2
= 40, 320

11. The number of permutations of n things
taken n at a time is given by:
nPn=
𝑛!
𝑛−𝑛 !
=
𝑛!
0!
= n!
CASE # 1:

12. Example #2:
Maia has 6 potted plants.
In how many ways can
she arrange these plants
in a row?

13. Given: n= 6 books
nPn = 6!
6P6 = (6)(5)(4)(3)(2)(1)
6P6 = 720 ways

14. The number of permutations of n things
taken r at a time is given by:
nPr=
𝑛!
𝑛−𝑟 !
!
CASE # 2:
n= no. of objects
r= no, of position

15. How many four-digit numbers
can be formed from the
numbers 1, 3, 4, 6, 8,
and 9 if repetition of digits is not
allowed?

16. Solution: There are four position to be
filled up.
How many choices are there for the first blank?
____________ x ____________ x ____________ x _____________

17. Solution: There are four position to be
filled up.
_____6______ x ____________ x ____________ x _____________
How many choices are there for the second blank?

18. Solution: There are four position to be
filled up.
_____6______ x _____5_____ x ____________ x _____________
How many choices are there for the third blank?

19. Solution: There are four position to be
filled up.
_____6______ x _____5_____ x _____4____ x _____________
How many choices are there for the fourth blank?

20. Solution: There are four position to be
filled up.
_____6______ x _____5_____ x _____4____ x _____3______
= 360
four- digit numbers will be formed

21. r
n P
=
𝟔!
𝟔−𝟒 !
=
𝟔!
𝟐 !
=
𝟔𝒙 𝟓 𝒙 𝟒 𝒙 𝟑 𝒙 𝟐 𝒙 𝟏
𝟐 𝒙 𝟏
=
𝟕𝟐𝟎
𝟐
= 360

22. Fifteen cars enter a race. In how
many ways can the trophies for
first, second and third place be
awarded?

23. Permutation with like Objects
(Distinguishable Permutation) is denoted
by:
P=
𝑛!
𝑛1!𝑛2!…𝑛𝑘!
!
CASE # 3:
n= no. of objects
nk = represent the
letters or object
that is repeated.

24. How many ways can the letters of the
word “TAGAYTAY” be arranged?

25. Solution:
Given: TAGAYTAY
Letters in all= 8
T’s= 2
A’s= 3
Y’s= 2
P=
𝑛!
𝑛1!𝑛2!..
P=
8!
2!3!2!
P=
40,320
24
P= 1680

26. In how many ways can the letters in the word
“TALLAHASSEE” be arranged?

27. TALLAHASSEE
Given:
“TALLAHASSEE”
Letters in all= 11
A’s= 3
L’s= 2
S’s= 2
E’s = 2
P=
𝑛!
𝑛1!𝑛2!..
P=
11!
3!2!2!2!
P=
39 916 800
48
P= 831, 600

28. Circular permutation
- is the total number of ways in which n distinct objects can be
arranged around a fixed circle. The number of ways to arrange n
distinct objects along a fixed circle is
P= (n-1)!!
CASE # 4:

29. In how many ways can 4 people be
seated around a circular table?

30. In how many ways can 5 boys and 4 girls
be seated around a circle table
without any restriction?

31. Permutation
WITH
restrictions

32. In how many ways can 3 boys and 4 girls
be seated in a table;
A. Without restriction
B. The parents stand together
C. All the females stand together

33. 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.

34. 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.

35. 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.

36. How many three-digit numbers greater than
300 may be formed using the
numbers 1, 2, 3, 7 and 9 if repetition of digit is
allowed?