Permutations and Combinations: When to Use Which

Permutations and combinations are where many FSc, A Level and university students lose easy marks, not because the formulas are hard, but because they pick the wrong one. The whole topic comes down to one question: does order matter? This guide explains that question clearly and then works through typical exam problems step by step.

Start With Factorials

The factorial of a positive integer \(n\) is

$$ n! = n \times (n - 1) \times (n - 2) \times \cdots \times 2 \times 1 $$

By definition, \(0! = 1\). For example, \(5! = 120\) and \(7! = 5040\).

The Fundamental Counting Principle

If one task can be done in \(m\) ways and a second task in \(n\) ways, both together can be done in \(m \times n\) ways. Every formula in this topic is built on this principle.

Permutations: Order Matters

A permutation is an arrangement. Changing the order gives a new permutation. The number of ways to arrange \(r\) objects chosen from \(n\) different objects is

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

Use permutations for: arranging people in a line, forming numbers or codes, assigning distinct positions (captain, vice-captain), seating, and arranging letters of a word.

Combinations: Order Does Not Matter

A combination is a selection. Choosing \(\{A, B, C\}\) is the same as choosing \(\{C, A, B\}\). The number of ways to choose \(r\) objects from \(n\) is

$$ {}^nC_r = \binom{n}{r} = \frac{n!}{r!\,(n - r)!} $$

Use combinations for: choosing a team or committee, selecting questions on an exam, picking cards, or choosing items from a menu.

The Link Between Them

$$ {}^nP_r = {}^nC_r \times r! $$

First choose the group, then arrange its \(r\) members in \(r!\) ways.

The Decision Test

Ask yourself: "If I swap two chosen items, do I get a different outcome?"

• Yes → order matters → permutation.
• No → order doesn't matter → combination.

Worked Example 1: Same Numbers, Different Questions

From 7 different books:

(a) In how many ways can 3 books be arranged on a shelf?

Arranging means order matters, so use permutations:

$$ {}^7P_3 = \frac{7!}{4!} = 7 \times 6 \times 5 = 210 $$

(b) In how many ways can 3 books be chosen to take on a trip?

Choosing means order doesn't matter, so use combinations:

$$ {}^7C_3 = \frac{7!}{3!\,4!} = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = \frac{210}{6} = 35 $$

Notice that \(210 = 35 \times 3!\). Each group of 3 books can be arranged in 6 ways, which is exactly why the permutation count is 6 times larger.

Worked Example 2: Assigning Positions

A society of 12 members must elect a president and a secretary. In how many ways can this be done?

Step 1: Is order important? Yes. Ali as president and Sara as secretary is different from Sara as president and Ali as secretary.

Step 2: Use permutations:

$$ {}^{12}P_2 = \frac{12!}{10!} = 12 \times 11 = 132 $$

Worked Example 3: Committee With Conditions

A committee of 5 is to be formed from 6 men and 5 women so that it has exactly 3 men and 2 women. How many committees are possible?

Step 1: A committee is a selection, so use combinations.

Step 2: Choose 3 men from 6:

$$ {}^6C_3 = \frac{6 \times 5 \times 4}{3 \times 2 \times 1} = 20 $$

Step 3: Choose 2 women from 5:

$$ {}^5C_2 = \frac{5 \times 4}{2 \times 1} = 10 $$

Step 4: Both choices happen together, so multiply (counting principle):

$$ 20 \times 10 = 200 $$

Worked Example 4: Letters With Repetition

How many different arrangements can be made from the letters of the word PAKISTAN?

Step 1: Count the letters: P, A, K, I, S, T, A, N, which is 8 letters, with A repeated twice.

Step 2: When some objects are identical, divide by the factorial of each repetition:

$$ \frac{n!}{p!\,q!\cdots} $$

Step 3: Apply it:

$$ \frac{8!}{2!} = \frac{40320}{2} = 20160 $$

Two More Useful Results

• Circular arrangements: \(n\) different objects around a circle can be arranged in \((n - 1)!\) ways, because rotations are not counted as new arrangements. For example, 6 people at a round table: \(5! = 120\) ways.
• Symmetry of combinations: \({}^nC_r = {}^nC_{n-r}\). Choosing 8 players to play from 10 is the same as choosing the 2 to leave out: \({}^{10}C_8 = {}^{10}C_2 = 45\).

Common Mistakes

• Using permutations for teams or committees. A team has no order unless positions are assigned.
• Adding instead of multiplying. Use "and" → multiply; "or" (separate cases) → add.
• Forgetting repeated letters when arranging words like MISSISSIPPI or PAKISTAN.
• Treating a circle like a line. Fix one person, then arrange the rest.
• Calculating huge factorials fully. Cancel first: \(\frac{12!}{10!} = 12 \times 11\).

Key Takeaways

• The key question is: does order matter? Yes → permutation; no → combination.
• \({}^nP_r = \frac{n!}{(n - r)!}\) and \({}^nC_r = \frac{n!}{r!\,(n - r)!}\), linked by \({}^nP_r = {}^nC_r \times r!\).
• Multiply for independent stages ("and"), add for separate cases ("or").
• Divide by factorials of repeated items, and use \((n - 1)!\) for circular arrangements.

If you are unsure whether a question needs a permutation or a combination, send it to Amathify on WhatsApp or through the "Ask a Question" form. Our experts and PhD scholars will explain the reasoning and solve it step by step.