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Mathematics

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Permutations and combinations are fundamental concepts in combinatorics, dealing with the arrangement and selection of objects from a set.

Factorials

Before diving into permutations and combinations, it's essential to understand factorials.
The factorial of a non-negative integer nn, denoted by n!n!, is the product of all positive integers less than or equal to nn.

n!=n×(n−1)×(n−2)×⋯×3×2×1n! = n \times (n-1) \times (n-2) \times \dots \times 3 \times 2 \times 1

By definition, 0!=10! = 1.

Example:
5!=5×4×3×2×1=1205! = 5 \times 4 \times 3 \times 2 \times 1 = 120

Permutations

Permutations refer to the arrangement of objects where the order matters. If you change the order of the objects, it's considered a different permutation.

1. Permutations of nn distinct objects taken rr at a time

The number of permutations of nn distinct objects taken rr at a time is given by the formula:

P(n,r)=n!(n−r)!P(n, r) = \frac{n!}{(n-r)!}

Example 1: Arranging a subset
How many ways can you arrange 3 different books on a shelf from a selection of 5 different books?

Here, n=5n=5 (total books) and r=3r=3 (books to arrange).

P(5,3)=5!(5−3)!=5!2!=5×4×3×2×12×1=5×4×3=60P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = \frac{5 \times 4 \times 3 \times 2 \times 1}{2 \times 1} = 5 \times 4 \times 3 = 60

There are 60 ways to arrange the books.

2. Permutations of nn distinct objects taken nn at a time

This is a special case where r=nr=n. The formula simplifies to:

P(n,n)=n!(n−n)!=n!0!=n!1=n!P(n, n) = \frac{n!}{(n-n)!} = \frac{n!}{0!} = \frac{n!}{1} = n!

Example 2: Arranging all objects
How many different ways can the letters of the word "MATH" be arranged?

Here, n=4n=4 (total letters) and r=4r=4 (all letters are used).

P(4,4)=4!=4×3×2×1=24P(4, 4) = 4! = 4 \times 3 \times 2 \times 1 = 24

There are 24 different arrangements.

3. Permutations with Repetition (Identical Objects)

If there are nn objects where n1n_1 are of one type, n2n_2 are of a second type, ..., nkn_k are of a kk-th type, and n1+n2+⋯+nk=nn_1 + n_2 + \dots + n_k = n, then the number of distinct permutations is:

n!n1!n2!…nk!\frac{n!}{n_1! n_2! \dots n_k!}

Example 3: Arranging letters with repetition
How many distinct arrangements can be made from the letters of the word "MISSISSIPPI"?

Here, n=11n=11 (total letters).
M: 1 (n1=1n_1=1)
I: 4 (n2=4n_2=4)
S: 4 (n3=4n_3=4)
P: 2 (n4=2n_4=2)

Number of distinct arrangements = 11!1!4!4!2!=39,916,8001×24×24×2=39,916,8001152=34,650\frac{11!}{1! 4! 4! 2!} = \frac{39,916,800}{1 \times 24 \times 24 \times 2} = \frac{39,916,800}{1152} = 34,650
There are 34,650 distinct arrangements.

Combinations

Combinations refer to the selection of objects where the order does not matter. If you select a group of objects, changing the order in which you picked them does not create a new combination.

1. Combinations of nn distinct objects taken rr at a time

The number of combinations of nn distinct objects taken rr at a time is given by the formula:

C(n,r)=(nr)=n!r!(n−r)!C(n, r) = \binom{n}{r} = \frac{n!}{r!(n-r)!}

Note that C(n,r)=P(n,r)r!C(n, r) = \frac{P(n, r)}{r!}. This shows that for every r!r! permutations, there is only 1 combination.

Example 4: Selecting a committee
A committee of 3 people is to be chosen from a group of 7 people. How many different committees can be formed?

Here, n=7n=7 (total people) and r=3r=3 (people to choose). The order of selection does not matter for a committee.

C(7,3)=7!3!(7−3)!=7!3!4!=7×6×5×4×3×2×1(3×2×1)(4×3×2×1)=7×6×53×2×1=7×5=35C(7, 3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} = \frac{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}{(3 \times 2 \times 1)(4 \times 3 \times 2 \times 1)} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 7 \times 5 = 35

There are 35 different committees that can be formed.

Example 5: Selecting items from a menu
A student needs to choose 2 subjects from a list of 4 optional subjects. How many ways can the student choose the subjects?

Here, n=4n=4 (total subjects) and r=2r=2 (subjects to choose). The order of choosing subjects does not matter.

C(4,2)=4!2!(4−2)!=4!2!2!=4×3×2×1(2×1)(2×1)=122=6C(4, 2) = \frac{4!}{2!(4-2)!} = \frac{4!}{2!2!} = \frac{4 \times 3 \times 2 \times 1}{(2 \times 1)(2 \times 1)} = \frac{12}{2} = 6

There are 6 ways to choose the subjects.

Key Difference: Order Matters vs. Order Does Not Matter

  • Permutations: Use when the order or arrangement of the selected items is important (e.g., arranging books, forming a password, assigning positions).
  • Combinations: Use when only the selection or group of items is important, and their order does not change the outcome (e.g., choosing a committee, selecting lottery numbers, picking cards).
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