Difference Between Permutation and Combination

TL;DR
Permutation vs Combination is whether order matters: a permutation counts arrangements (order matters), while a combination counts selections (order doesn't). This article covers both definitions, the nPr and nCr formulas derived from scratch, the relationship between them, and how to tell which one a problem needs.

What Is the Difference Between Permutation and Combination?

A permutation is an arrangement of objects in which the order matters; a combination is a selection of objects in which the order does not matter. Both start from the same act — choosing r objects from a set of n — but a permutation also cares how those r are arranged, and a combination does not.

Pick 2 letters from A,B,C. As combinations, there are three selections: AB, AC, BC — and BA is the same selection as AB. As permutations, there are six arrangements: AB, BA, AC, CA, BC, CB — because AB and BA are different orderings. Same objects, same act of choosing; the only difference is whether the order counts. That's why a permutation count is always at least as large as the matching combination count.

How Do You Know Which One to Use?

This is the question every student actually wants answered, and the test is one sentence: if reordering the chosen items creates a new, distinct outcome, use permutation; if reordering gives you the same outcome, use combination.

A few reliable signal words help, though the order-test always wins over keyword-spotting:

Choosing a President, Vice-President, and Secretary from 10 people is a permutation — the three roles are distinct, so order matters. Choosing a 3-person committee from the same 10 is a combination — the three are interchangeable, so order doesn't. Same 10 people, same "choose 3," opposite tools.

The Formulas — Derived, Not Just Stated

Both formulas come from one idea: counting step by step, then correcting for what you over-counted. Start with the factorial.

Factorial. n! is the product of all positive integers up to n: 5!=5×4×3×2×1=120. It counts the ways to arrange n distinct objects in a row, and by convention 0!=1.

Permutation formula. To arrange r objects chosen from n, fill r slots in order. The first slot has n choices, the second n−1, down to n−r+1 for the last. Multiply:

[ nPr=n×(n−1)×⋯×(n−r+1)=\frac{n!}{(n−r)!} ]

The ( (n−r)! ) in the denominator cancels the tail of the product you didn't use. For 5P2=5!3!=120/6=20.

Combination formula. A combination is a permutation that stopped caring about order. Each selection of r objects can be arranged in r! different orders, and a permutation counts all of them — so to count selections, divide the permutation count by r!:

[ nCr=\frac{nPr}{r!} = \frac{n!}{r!(n−r)!} ]

For 5C2=5!/(2!3!)=120/(2×6)=10 — exactly half of 5P2=20, because each pair has 2! orderings.

The relationship. Reading the combination derivation backward gives the link between the two:

[ nPr=nCr×r! ]

A permutation is a combination times the number of ways to order what you chose. That single equation is the difference between permutation and combination, written in symbols.

Examples of the Difference Between Permutation and Combination

Example 1
In how many ways can a President and a Vice-President be chosen from 6 club members?
Two distinct roles, so order matters — a permutation. Fill the President slot (6 choices), then VP (5 choices):
[ 6P2=\frac{6!}{4!} = 6 × 5 = 30 ]
Final answer: 30 ways.

Example 2
From 10 people, how many ways can a 3-person committee be formed?
Wrong attempt. A student reasons "10 choices for the first member, 9 for the second, 8 for the third" and writes 10×9×8=720. The check: a committee of {Asha, Ben, Carl} is the same committee however it's listed, but 720 counts Asha-Ben-Carl, Ben-Carl-Asha, and the other four orderings as separate — six times each. The count is inflated by exactly 3!.
Correct. Because order doesn't matter in a committee, this is a combination. Divide the permutation count by 3!:

[ 10C3=\frac{10!}{3!7!} = \frac{720}{6} = 120 ]
Final answer: 120 committees.

Example 3
How many 4-letter arrangements can be made from the letters of "MATH" (no repeats)?
All four letters used, all distinct, order matters — a permutation of 4 objects: [ 4P4=4! = 24 ]
Final answer: 24 arrangements.

Example 4
How many 3-card hands can be dealt from a standard 52-card deck?
A hand is a selection — the order the cards arrive doesn't change the hand — so it's a combination:
[ 52C3=\frac{52!}{3!49!} = \frac{52×51×50}{6} = 22,100 ]
Final answer: 22,100 hands.

Example 5
A 4-digit PIN uses digits 0–9 with no repeats. How many PINs are possible?
Order matters in a PIN — 1234 and 4321 open different locks — and digits don't repeat, so it's a permutation of 4 from 10: [ 10P4=\frac{10!}{6!} = 10×9×8×7 = 5,040 ]
Final answer: 5,040 PINs.

Example 6
From 7 men and 5 women, how many committees of 2 men and 2 women can be formed?
Both choices are selections (committee, no roles), so both are combinations — and because the two choices happen together, multiply them: [ 7C2×5C2=\frac{7×6}{2}×\frac{5×4}{2}=210 ]
Final answer: 210 committees.

Why the Distinction Matters Beyond the Classroom

Order-matters-or-not is not a textbook nicety. It decides how big a space of possibilities actually is, and that drives real systems.

• Security. A 4-digit PIN is far harder to guess than an unordered selection would be — the strength of a passcode is its permutation count.

• Probability and cards. Poker hand odds are combination counts; the chance of a flush is a ratio of nCr values.

• Genetics and sampling. Choosing a sample of individuals from a population is a combination; the number of possible samples is an nCr.

• Scheduling. Arranging tasks or speakers in a sequence is a permutation; the number of possible schedules is a factorial-based nPr.

Tripping Points to Avoid

Almost every error here is a misread of whether order matters. Each fix routes back to the one-sentence test.

Mistake 1: Using permutation when order doesn't matter

Where it slips in: Committee, team, or "any r of them" problems, where the rusher counts n×(n−1)×⋯ out of habit.

Don't do this: Treat a committee selection as a permutation — 10×9×8=720 for a 3-person committee.

The correct way: Order doesn't matter in a committee, so it's a combination — divide by r!: 10C3=720/6=120.

Mistake 2: Forgetting that nPr=nCr×r!

Where it slips in: Switching between the two formulas mid-problem, where the second-guesser recomputes from scratch instead of converting.

Don't do this: Treat permutations and combinations as unrelated, deriving each separately every time.

The correct way: They're one idea apart — nPr=nCr×r!. If you have one, you have the other by multiplying or dividing by r!.

Mistake 3: Mishandling repeats or restrictions

Where it slips in: Problems with repeated elements or "no repeats" rules, where the memorizer applies the plain formula blindly.

Don't do this: Use nPr for arrangements of objects that aren't all distinct, or ignore a "without repetition" condition.

The correct way: Read the conditions first. Repeated objects need the permutations-with-repetition formula; "no repeats" means standard nPr. The formula is the last step, not the first.

Key Takeaways

• The difference between permutation and combination is order: permutations count arrangements (order matters), combinations count selections (order doesn't).

• The one-sentence test: if swapping two chosen items makes a new outcome, it's a permutation; if not, a combination.

• The formulas are nPr=n!(n−r)! and nCr=n!/(r!(n−r)!).

• They are linked by nPr=nCr×r! — a permutation is a combination times the orderings of what you chose.

• Permutation counts are always at least as large as the matching combination counts.