Variations in Counting
1. Arranging a Set with Non-Distinct (Repeated) Elements
When arranging objects that contain repeated or identical elements, some arrangements look the same because the identical elements cannot be distinguished from one another.
To find the number of distinct arrangements, we use the multinomial permutation formula.
n! / (n1! n2! ….. nk!)
If a set contains:
n = total elements,
n1 = identical elements of one type,
n2 = identical elements of another type,
and so on, then the number of distinct arrangements is:
Example 1.
How many distinct arrangements can be made from the letters in LEVEL?
Letters:
L = 2
E = 2
V = 1
Total letters = 5
n! / (n1! n2! ….. nk!)
= 5! / (2!2!1!)
= (5x4x3x2x1) / ((2x1) (2x1) (1))
= 120/4
= 30 arrangements
Example 2.
How many distinct arrangements can be made from the letters in BALLOON?
Letters:
B = 1
A = 1
L = 2
O = 2
N = 1
Total letters = 7
n! / (n1! n2! ….. nk!)
= 7! / (1!1!2!2!1!)
= (7x6x5x4x3x2x1) / ((1) (1) (2x1) (2x1) (1))
= 5040 / 4
= 1260 arrangements
Example 3.
A teacher has 7 books to arrange on a shelf: 3 copies of Math, 2 copies of Science, and 2 copies of English.
In how many distinct ways can they be arranged?
Books:
Math = 3
Science = 2
English = 2
Total books = 7
n! / (n1! n2! ….. nk!)
= 7! / (3!2!2!)
= (7x6x5x4x3x2x1) / ((3x2x1) (2x1) (2x1))
= 5040 / 24
= 210 arrangements
Example 4.
A gift box contains 6 candies: 2 caramel, 2 mint, and 2 strawberry. How many visually different candy arrangements are possible?
Candies:
caramel = 2
mint = 2
strawberry = 2
Total candies = 6
n! / (n1! n2! ….. nk!)
= 6! / (3!2!2!)
= (6x5x4x3x2x1) / ((2x1) (2x1) (2x1))
= 720 / 8
= 90 arrangements
2. Arranging objects in a circle (circular permutations)
When objects are arranged in a line, positions are distinct. In a circle, rotations are considered the same arrangement — so we “fix” one object to remove rotational symmetry.
For n distinct objects in a circle:
(n−1)!
This is the number of circular permutations.
Example 1.
Arrange 5 people around a round table.
(n−1)!
= (5−1)!
= (4)!
= 4x3x2x1
= 24 seating's
Example 2.
A bracelet is made with 5 different colored beads. How many different circular arrangements are possible?
(n−1)!
= (5−1)!
= (4)!
= 4x3x2x1
= 24 arrangements