Permutation and Combination

Definition

Permutation is an arrangement of objects in a specific order. If the order of the selected objects matters, the counting is done using permutations.

For $n$ distinct objects taken $r$ at a time, the number of permutations is:

$${}^{n}P_{r} = \frac{n!}{(n-r)!}$$

Combination is a selection of objects without regard to order. If the order of the selected objects does not matter, the counting is done using combinations.

For $n$ distinct objects taken $r$ at a time, the number of combinations is:

$${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$$

Where:

• $n!$ means factorial of $n$
• $n! = n \times (n-1) \times (n-2) \times \cdots \times 2 \times 1$
• $0! = 1$

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Main Content

1. Permutation

Order matters

• In permutation, changing the order creates a different arrangement.

Examples

• Arranging books on a shelf, seating people in chairs, forming PIN codes, selecting president/secretary roles from a group.

If we choose $r$ objects from $n$ distinct objects and arrange them, the number of ways is:

$${}^{n}P_{r} = \frac{n!}{(n-r)!}$$

Example 1: Arranging 3 people from 5

Suppose 5 people are available and we need to arrange 3 of them in a row.

$${}^{5}P_{3} = \frac{5!}{(5-3)!} = \frac{120}{2} = 60$$

So, there are 60 different arrangements.

Example 2: Using all objects

If 4 different letters $A, B, C, D$ are arranged, then:

$${}^{4}P_{4} = 4! = 24$$

The arrangements include: $$ABCD, ABDC, ACBD, ACDB, ADBC, ADCB, \dots$$

Key idea

Permutation counts different orderings of the same set of selected objects.
For example:

• $AB$ and $BA$ are different permutations.
• $AB$ and $BA$ are the same combination.

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2. Combination

Order does not matter

• In combination, only the chosen objects matter, not the sequence in which they are selected.

Examples

• Forming a team, choosing a group of students, selecting lottery numbers, deciding ingredients for a recipe.

If we choose $r$ objects from $n$ distinct objects, the number of combinations is:

$${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$$

Example 1: Choosing 3 students from 5

Suppose 5 students are available and we want to choose a group of 3.

$${}^{5}C_{3} = \frac{5!}{3!2!} = \frac{120}{6 \times 2} = 10$$

So, there are 10 possible groups.

Example 2: Choosing 2 items from 4

If the items are $A, B, C, D$, the combinations of 2 are:

$$AB, AC, AD, BC, BD, CD$$

Notice:

• $AB$ and $BA$ are not counted separately.
• This is because the group $\{A, B\}$ is the same regardless of order.

Key idea

Combination counts selections, not arrangements.

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3. Relationship Between Permutation and Combination

Permutation is based on combination

• A permutation can be formed by first selecting objects and then arranging them.

Formula relationship

$${}^{n}P_{r} = {}^{n}C_{r} \times r!$$

This means:

1. First choose $r$ objects from $n$
2. Then arrange those $r$ chosen objects in all possible orders

Why this works

Once a group of $r$ objects is selected, they can be arranged in $r!$ ways.

So:

$${}^{n}C_{r} = \frac{{}^{n}P_{r}}{r!}$$

Example

For 5 objects taken 3 at a time:

$${}^{5}P_{3} = 60$$

$${}^{5}C_{3} = 10$$

And:

$${}^{5}C_{3} \times 3! = 10 \times 6 = 60$$

This confirms:

$${}^{n}P_{r} = {}^{n}C_{r} \times r!$$

Visual idea for order vs no order

Selected objects: A, B, C

Permutation:
ABC, ACB, BAC, BCA, CAB, CBA

Combination:
{A, B, C}

The permutation list has 6 arrangements, while the combination has only 1 selection.

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Working / Process

1. Read the problem carefully and identify whether order matters

• If rearrangement, ranking, seating, or assigning positions is involved, use permutation.
• If only choosing a group or subset is involved, use combination.

2. Determine the values of $n$ and $r$

• $n$ = total available objects
• $r$ = number of objects selected or arranged

3. Apply the correct formula and simplify

• Use permutation formula when order matters: $${}^{n}P_{r} = \frac{n!}{(n-r)!}$$

• Use combination formula when order does not matter: $${}^{n}C_{r} = \frac{n!}{r!(n-r)!}$$

• Compute factorials carefully and reduce fractions to avoid errors.

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Advantages / Applications

Solves counting problems efficiently

• Permutation and combination provide systematic methods to count possibilities without listing them all.

Useful in real-life decision-making

• They are applied in forming committees, scheduling, ranking candidates, designing codes, and selecting teams.

Foundation for probability and discrete mathematics

• These concepts are essential for calculating probabilities, analyzing algorithms, and studying combinatorial structures.

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Summary

• Permutation is used for arrangement, while combination is used for selection.
• Order matters in permutation and does not matter in combination.
• The two are connected by the relation ${}^{n}P_{r} = {}^{n}C_{r} \times r!$.
• Important terms to remember: factorial, arrangement, selection, order matters, order does not matter.