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. $n!$ ("$n$ factorial") is the product of all positive integers up to $n$: $5! = 5 \times 4 \times 3 \times 2 \times 1 = 120$. 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:
$$nPr = n \times (n-1) \times \cdots \times (n - r + 1) = \frac{n!}{(n - r)!}$$
The $(n-r)!$ in the denominator cancels the tail of the product you didn't use. For $5P2 = \frac{5!}{3!} = \frac{120}{6} = 20$.
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 $5C2 = \frac{5!}{2! , 3!} = \frac{120}{2 \times 6} = 10$ — exactly half of $5P2 = 20$, because each pair has $2! = 2$ orderings.
The relationship. Reading the combination derivation backward gives the link between the two:
$$nPr = nCr \times r!$$
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
The examples move from a clean count to a real over-counting trap to a problem that uses both tools at once.
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):
$$6P2 = \frac{6!}{4!} = 6 \times 5 = 30$$
Final answer: 30 ways. President-then-VP is a different outcome from VP-then-President, which is why each pairing is counted twice over — correctly, since the roles differ.
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 $10 \times 9 \times 8 = 720$. 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 $3! = 6$.
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. The $720$ was the number of ordered selections; dividing by $3!$ collapses the six orderings of each committee into one.
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. With every object placed and order counting, the answer is just the factorial.
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:
$$52C3 = \frac{52!}{3! , 49!} = \frac{52 \times 51 \times 50}{6} = \frac{132600}{6} = 22100$$
Final answer: 22,100 hands. Had the order mattered (dealing to three distinct positions), it would be $52P3 = 132600$ — exactly $3!$ times larger.
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:
$$10P4 = \frac{10!}{6!} = 10 \times 9 \times 8 \times 7 = 5040$$
Final answer: 5,040 PINs. This is why a "combination lock" is misnamed; the math behind it is a permutation.
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:
$$7C2 \times 5C2 = \frac{7 \times 6}{2} \times \frac{5 \times 4}{2} = 21 \times 10 = 210$$
Final answer: 210 committees. The multiplication principle joins two independent combination counts — a standard step up from single-tool problems.
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 ($10P4$ with order) 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. Order doesn't matter to a hand, so combinations are the right tool.
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 $nPr$.
The destination this points to is the binomial theorem and probability — the combination count $nCr$ is exactly the binomial coefficient $\binom{n}{r}$ that builds Pascal's triangle and powers the whole probability curriculum. A reader will meet these as "$n$ choose $r$" the moment they open a probability chapter.
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 $n \times (n-1) \times \cdots$ out of habit.
Don't do this: Treat a committee selection as a permutation — $10 \times 9 \times 8 = 720$ 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 $nPr = nCr \times r!$
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 — $nPr = nCr \times r!$. 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 (letters in "MISSISSIPPI") 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 $\frac{n!}{n_1! , n_2! \cdots}$; "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 $nPr = \dfrac{n!}{(n-r)!}$ and $nCr = \dfrac{n!}{r!,(n-r)!}$.
They are linked by $nPr = nCr \times r!$ — 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.
Practice These Three Before Moving On
From 8 runners, how many ways can gold, silver, and bronze be awarded?
From 8 runners, how many ways can a 3-person relay team be chosen (no positions)?
Explain in one line why your answer to Problem 1 is exactly $3!$ times your answer to Problem 2.
If your answers to Problems 1 and 2 are equal, return to Mistake 1 — one of them must account for order.
The same "choose then arrange" thinking underpins how a set lists its elements, how subsets are counted, and how an intersection of sets selects shared members. Want a live Bhanzu trainer to walk your child through permutations, combinations, and how to tell them apart fast? Book a free demo class — online globally.
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