The permutations and combinations formula answers one of the most common questions in counting: in how many ways can you pick or arrange items from a larger group? The only difference between the two is whether order matters. If it does, you use permutations (nPr). If it does not, you use combinations (nCr).
Picking the wrong one is the most frequent mistake in probability homework, lottery odds, password strength estimates and scheduling problems. This guide shows both formulas, how to decide between them, how to work each one out by hand, and how to check your answer with a calculator on Calculatorr.
What is the difference between permutations and combinations?
A permutation is an ordered arrangement. A combination is an unordered selection. The same items can give very different counts depending on which one the problem calls for.
Take three people, Ana, Ben and Carla, and pick two of them:
- As a permutation (first place and second place): Ana-Ben, Ben-Ana, Ana-Carla, Carla-Ana, Ben-Carla, Carla-Ben. That is 6 results.
- As a combination (a team of two): Ana-Ben, Ana-Carla, Ben-Carla. That is 3 results.
Each team of two can be ordered in 2 ways, which is why the permutation count is exactly twice the combination count here.
Quick test to decide which one to use
Ask yourself: if I swap two selected items, do I get a different outcome?
- Yes: use permutations. Examples: podium positions, PIN codes, seating plans, race rankings, president and vice president.
- No: use combinations. Examples: committees, lottery numbers, card hands, pizza toppings, choosing teammates.
Permutations formula: how to find nPr
The number of ways to arrange r items chosen from n distinct items is:
nPr = n! / (n − r)!
Here n is the total number of items, r is how many you select, and the exclamation mark means factorial: the product of all whole numbers from that number down to 1. By definition, 0! = 1.
Step-by-step method for nPr
- Identify n (total items) and r (items chosen).
- Check that r is not larger than n.
- Write n! / (n − r)!.
- Cancel the common factors instead of computing full factorials.
- Multiply the remaining numbers.
A useful shortcut: nPr equals the product of r numbers starting at n and counting down. So 10P3 = 10 × 9 × 8.
Example: podium finishes in a race
Ten runners compete and you want the number of possible gold, silver and bronze outcomes.
- n = 10, r = 3
- 10P3 = 10! / 7! = 10 × 9 × 8 = 720
There are 720 different ways the top three places can be filled.
Example: a PIN with no repeated digits
How many 4-digit PINs can you build from digits 0 to 9 if no digit repeats?
- n = 10, r = 4
- 10P4 = 10 × 9 × 8 × 7 = 5,040
If digits are allowed to repeat, the count changes to 10 × 10 × 10 × 10 = 10,000, which is covered further below.
Combinations formula: how to find nCr
The number of ways to choose r items from n distinct items, ignoring order, is:
nCr = n! / (r! × (n − r)!)
It is the permutation formula divided by r!, which removes the duplicate orderings of each group. That gives the key relationship:
nPr = nCr × r!
Step-by-step method for nCr
- Identify n and r.
- Compute the numerator as the product of r numbers counting down from n.
- Compute the denominator as r!.
- Divide.
Example: choosing a committee
A club has 10 members and needs a 3-person committee with no roles assigned.
- 10C3 = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120
Notice that 720 is the same number as 10P3. Dividing by 3! = 6 removes the repeated orderings of each trio.
Example: pizza toppings
You can choose 3 toppings from a menu of 8 and the order you pick them does not matter.
- 8C3 = (8 × 7 × 6) / 6 = 336 / 6 = 56
There are 56 different three-topping pizzas.
nPr vs nCr comparison table
This table shows how quickly the two counts diverge for the same n and r.
| n | r | nPr (order matters) | nCr (order ignored) |
|---|---|---|---|
| 5 | 2 | 20 | 10 |
| 6 | 3 | 120 | 20 |
| 8 | 3 | 336 | 56 |
| 10 | 3 | 720 | 120 |
| 10 | 4 | 5,040 | 210 |
Each nCr value equals the matching nPr value divided by r!. For r = 3 that divisor is 6, and for r = 4 it is 24.
Useful rules and shortcuts for nCr and nPr
- 0! = 1. This makes nC0 = 1 and nCn = 1, since there is exactly one way to choose nothing or everything.
- Symmetry: nCr = nC(n − r). So 49C43 equals 49C6. Always use the smaller of r and n − r to reduce work.
- nP0 = 1 and nPn = n!. Arranging all n items gives n! orderings.
- Pascal's triangle: nCr = (n−1)C(r−1) + (n−1)Cr. Each entry is the sum of the two above it.
How to find the odds of a lottery with the combinations formula
Lottery draws are a classic combinations problem because the order of the drawn numbers does not matter. Suppose you pick 6 numbers from 49.
- 49C6 = (49 × 48 × 47 × 46 × 45 × 44) / (6 × 5 × 4 × 3 × 2 × 1)
- Numerator = 10,068,347,520
- Denominator = 720
- Result = 13,983,816
Your chance of matching all six numbers with one ticket is 1 in 13,983,816, or about 0.0000072%.
The same method applies to card games. A five-card poker hand from a 52-card deck gives 52C5 = 2,598,960 possible hands.
Counting with restrictions: combining several choices
Many real problems split into separate selections. Multiply the combinations for each group.
Example: a project team needs 4 women from a pool of 6 and 3 men from a pool of 5.
- Women: 6C4 = 15
- Men: 5C3 = 10
- Total teams: 15 × 10 = 150
For "at least" problems, it is often easier to count the total and subtract the cases you do not want. For example, teams of 3 from 10 people with at least one specific person excluded: total 10C3 = 120, teams without a person X = 9C3 = 84, so teams that include X = 120 − 84 = 36.
Permutations with repetition and identical items
The basic formulas assume every item is different and can be used once. Two common variations change the formula.
When items can repeat: n to the power r
If each of r positions can be filled by any of n options, the count is nr. A 4-digit PIN where digits may repeat has 104 = 10,000 possibilities.
When some items are identical
To arrange n items where some are identical, divide n! by the factorial of each repeated group:
n! / (a! × b! × ...)
Example: how many distinct arrangements of the letters in LEVEL? There are 5 letters, with L twice and E twice.
- 5! / (2! × 2!) = 120 / 4 = 30
Circular arrangements
When items sit around a round table and rotations count as the same arrangement, the count is (n − 1)!. Six people around a table can be seated in 5! = 120 ways.
Combinations with repetition
If you can choose the same item more than once and order does not matter, use (n + r − 1)Cr. Choosing 3 scoops from 5 ice cream flavors, repeats allowed, gives 7C3 = 35 possibilities.
How to use an nPr and nCr calculator online
A calculator is the fastest way to confirm a result, especially when factorials get large. 20! already has 19 digits, and most basic calculators overflow quickly.
- Enter the total number of items (n).
- Enter the number of items selected (r).
- Choose permutation (nPr) or combination (nCr).
- Read the result and compare it with your hand estimate.
You can find this and other tools on Calculatorr. If you are working through related math topics, the same site includes calculators for fractions, percentages and statistics that pair well with counting problems.
How to interpret nPr and nCr results
The result is a count of distinct outcomes, not a probability. To turn it into a probability, divide the number of favorable outcomes by the total number of possible outcomes, using the same counting method for both.
Example: what is the chance that a random 3-person committee from 10 people includes both Ana and Ben?
- Total committees: 10C3 = 120
- Committees with Ana and Ben: the third member comes from the other 8 people, so 8 committees
- Probability = 8 / 120 = 1/15, about 6.67%
As a rule of thumb, a very large count means a very large space of possibilities. That is why long passwords with many character options are hard to guess, and why lottery odds are so low.
Common mistakes when using the permutations and combinations formula
- Using nCr when order matters. If positions or roles differ, you need nPr.
- Forgetting to divide by r!. This turns a combination into a permutation by accident.
- Setting r larger than n. Without repetition, you cannot select more items than exist, so the result is 0 or undefined.
- Computing full factorials. Cancel terms first. 49C6 never requires calculating 49!.
- Ignoring identical items. Arranging letters or objects that repeat requires dividing by the factorial of each repeated group.
- Mixing up n and r. n is the full pool and r is the number you pick. Reading the problem twice avoids this.
- Adding instead of multiplying. Independent choices multiply. Alternative cases that cannot happen together add.
Practice problems with answers
- How many ways can 5 books be arranged on a shelf from a collection of 8 distinct books? 8P5 = 8 × 7 × 6 × 5 × 4 = 6,720
- How many 2-card hands can be dealt from a 52-card deck? 52C2 = (52 × 51) / 2 = 1,326
- In how many ways can a manager choose 4 employees from 12 for a training group? 12C4 = (12 × 11 × 10 × 9) / 24 = 495
- How many distinct arrangements exist for the letters in BANANA? 6! / (3! × 2!) = 720 / 12 = 60
Try solving each by hand first, then verify with the calculator. Working both ways builds the habit of recognizing whether a problem is about ordering or selecting, which is the skill that matters most.
