# 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:

| Use **permutation** when the problem is about… | Use **combination** when the problem is about… |  
| --- | --- |  
| arranging, ordering, ranking | selecting, choosing, picking |  
| seating, lining up, sequencing | forming a team, group, or committee |  
| passwords, PINs, codes | a hand of cards, a subset, a sample |  
| 1st / 2nd / 3rd place (distinct roles) | "any 3 of them" (no roles) |

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.