Permutations & Combinations: nPr, nCr Formulas + Examples

Permutations and combinations are two ways of counting how many outcomes a selection can produce. The single question that separates them is simple: does the order of the chosen items matter? If swapping two items gives a genuinely different result, you are counting permutations. If a swap leaves you with the same group, you are counting combinations.

Both are built on the factorial, where $n! = n \times (n-1) \times \cdots \times 2 \times 1$ and $0! = 1$ by convention.

The two formulas

For choosing $r$ items from $n$ distinct items:

{}^{n}P_{r} = \frac{n!}{(n-r)!}, \qquad {}^{n}C_{r} = \binom{n}{r} = \frac{n!}{r!\,(n-r)!}

The permutation count ${}^{n}P_{r}$ treats two arrangements with the same members but different order as distinct. The combination count ${}^{n}C_{r}$ treats them as the same. The only difference between the formulas is the extra $r!$ in the denominator of ${}^{n}C_{r}$ — and that single factor carries the entire "order matters / order doesn't" distinction.

When to use which

Use this decision rule before reaching for either formula:

Situation	Order matters?	Use
Ranking finishers 1st/2nd/3rd in a race	Yes	${}^{n}P_{r}$
Assigning distinct roles (president, treasurer)	Yes	${}^{n}P_{r}$
Forming a PIN or password from distinct digits	Yes	${}^{n}P_{r}$
Choosing a committee with equal members	No	${}^{n}C_{r}$
Dealing a hand of cards	No	${}^{n}C_{r}$
Picking lottery numbers	No	${}^{n}C_{r}$

A quick test: rename the chosen items as "Group A." If a different seating, ranking, or labeling of the same people changes the answer, order matters → permutation. If only who is in Group A matters, order does not → combination.

Why the combination formula has that extra $r!$ — a derivation

Start from permutations. To arrange $r$ items chosen from $n$ , fill $r$ slots in turn: the first slot has $n$ choices, the next $n-1$ , down to $n-r+1$ choices for the last slot. Multiplying gives

{}^{n}P_{r} = n (n-1)\cdots (n-r+1) = \frac{n!}{(n-r)!}

Now notice that every combination of $r$ items can itself be arranged internally in $r!$ different orders, and each of those orders is a separate permutation. So the permutations are exactly the combinations counted $r!$ times over:

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

Dividing both sides by $r!$ isolates the combination count:

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

That is the whole story: a combination is a permutation with the internal orderings collapsed, and dividing by $r!$ does the collapsing.

Worked example 1 (permutation): a race podium

Eight runners compete. How many distinct ways can the gold, silver, and bronze medals be awarded?

Order matters — gold for runner A and silver for runner B is different from the reverse. So use ${}^{8}P_{3}$ :

{}^{8}P_{3} = \frac{8!}{(8-3)!} = \frac{8!}{5!} = 8 \times 7 \times 6 = 336

There are 336 possible podiums. The slot-filling logic confirms it directly: $8$ choices for gold, $7$ for silver, $6$ for bronze.

Worked example 2 (combination): a committee

From the same $8$ people, how many ways can a committee of $3$ (all equal members) be formed?

Order does not matter — a committee of {A, B, C} is identical to {C, B, A}. Use ${}^{8}C_{3}$ :

{}^{8}C_{3} = \frac{8!}{3!\,(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56

There are 56 committees. Notice the answer is exactly the permutation result $336$ divided by $3! = 6$ , which is the $r!$ factor in action: each committee was over-counted $6$ times by the permutation, once for each ordering of its three members.

Practice

Try these before checking the answers below.

A password uses $4$ different digits from $0$ – $9$ , and order matters. How many are possible?
A pizza shop has $7$ toppings; you pick $3$ different ones. How many topping sets?
In how many ways can $5$ distinct books be arranged on a shelf?

Answers: (1) ${}^{10}P_{4} = \dfrac{10!}{6!} = 5040$ . (2) ${}^{7}C_{3} = \dfrac{7!}{3!\,4!} = 35$ . (3) ${}^{5}P_{5} = 5! = 120$ .

Calculation traps to avoid

Choosing the wrong formula. The most common error is using ${}^{n}C_{r}$ when order matters (or vice versa). Always run the "does a swap change the answer?" test first.
Forgetting $0! = 1$ . When $r = n$ , the denominator includes $(n-r)! = 0! = 1$ , not $0$ . Setting $0! = 0$ breaks the formula.
Mixing up $n$ and $r$ . The total pool is $n$ ; the number chosen is $r$ , with $r \le n$ . ${}^{8}P_{3}$ is not the same as ${}^{3}P_{8}$ (the latter is undefined).
Cancelling factorials carelessly. $\frac{8!}{5!}$ is $8\times 7\times 6$ , not $\frac{8}{5}$ . Expand only down to where the smaller factorial cancels.
Ignoring repeated items. When some items are identical (like the three A's in "BANANA"), divide the total permutations by the factorial of each repeat group, or you will over-count distinct arrangements.

The fastest way to stop confusing these two is to internalize the relationship ${}^{n}P_{r} = {}^{n}C_{r}\times r!$ — every combination question is just a permutation question with the ordering divided back out.

Frequently Asked Questions

What is the difference between a permutation and a combination?: A permutation counts arrangements where order matters, so swapping two items gives a different outcome. A combination counts selections where order does not matter. The formulas differ only by an extra factor of r! in the denominator of the combination.
When should I use nPr instead of nCr?: Use nPr when order matters, such as ranking finishers, assigning distinct roles, or building a PIN from distinct digits. Use nCr when only the group membership matters, such as choosing a committee, dealing a hand, or picking lottery numbers.
How are nPr and nCr related?: They satisfy nPr = nCr times r!. Every combination can be internally arranged in r! orders, and each order is a separate permutation, so dividing the permutation count by r! gives the combination count.
Why is 0! equal to 1?: Defining 0! = 1 keeps the formulas consistent. For example, when r = n the denominator contains (n - r)! = 0!, and only 0! = 1 makes nCn equal to 1, which is the correct count of choosing all items in exactly one way.
How do I count arrangements when some items are identical?: Take the total factorial of all items and divide by the factorial of each repeated group. For the word BANANA with three A and two N, the count is 6! / (3! 2!) = 60 distinct arrangements.

Need help with a problem?

Upload your question and get a verified, step-by-step solution in seconds.

Open GPAI Solver →