Combination and Permutation Calculator
Values
Fill in the values and click Calculate to see results
Example 1
A lottery draws 6 distinct numbers from 1 to 49. How many different tickets exist, and what is the probability that a single ticket matches all six?
Solution
Order does not matter – a ticket holds a set of numbers – and numbers cannot repeat, so this is a combination without repetition:
Working the multiplicative form:
The numerator multiplies to and the denominator to 720, giving
Answer
There are 13,983,816 possible tickets, so a single ticket wins with probability 1 in 13,983,816 – roughly one jackpot per fourteen million tickets sold, the figure printed on every 6/49 play slip.
Source: Rosen, Discrete Mathematics and Its Applications
Example 2
An ice cream parlour offers 10 flavours and sells cones of exactly 3 scoops. Scoops stack, so repetition is allowed and order does not matter – vanilla-chocolate-vanilla is the same cone however the scoops are sequenced. How many different cones can the shop list?
Solution
Repetition allowed, order irrelevant: the with-repetition combination applies.
Expanding the binomial:
So the menu holds
Answer
The parlour can list 220 distinct triple-scoop cones – far more than the 120 available if every scoop had to be a different flavour, which is the repetition premium in dessert form.
Source: OpenStax Introductory Statistics, counting chapter
Example 3
A standard deck of 52 cards is shuffled. How many distinct orderings of the full deck exist?
Solution
An ordering of all 52 cards is a permutation of everything – the factorial:
Working through the product:
For scale: the universe is about 4.3 × 10¹⁷ seconds old, so a shuffled deck every second since the Big Bang would not reach even a billionth of a billionth of the total.
Answer
A shuffled deck has 52! ≈ 8.07 × 10⁶⁷ distinct orderings – which is why no well-shuffled deck in history has ever repeated an order.
Source: Graham, Knuth & Patashnik, Concrete Mathematics
Example 4
A system requires 5-character passwords using the 62 letters and digits. Characters may repeat. How many passwords exist, and how many more would 8 characters add?
Solution
Each of 5 slots takes any of 62 characters independently – a with-repetition permutation:
Evaluating the power:
At 8 characters the space grows to
Answer
Five characters give 916,132,832 passwords; eight give over 218 trillion – each extra character multiplies the space by 62, which is the entire argument for longer passwords.
Source: NIST digital identity guidance background
Published:
Updated:
References
- [1]NIST, NIST/SEMATECH e-Handbook of Statistical Methods. https://www.itl.nist.gov/div898/handbook
- [2]Khan Academy, Statistics and probability: counting, permutations, and combinations. https://www.khanacademy.org/math/statistics-probability/counting-permutations-and-combinations
- [3]McGraw-Hill, Discrete Mathematics and Its Applications, 2018.
- [4]Addison-Wesley, Concrete Mathematics: A Foundation for Computer Science, 1994.
- [5]OpenStax, Rice University, OpenStax Introductory Statistics. https://openstax.org/details/introductory-statistics
Glossary
- Combination – A selection of k items from a pool of n where the order of selection does not matter, counted by C(n, k) = n!/[k!(n−k)!].
- Permutation – An arrangement of k items from a pool of n where order is part of the outcome, counted by P(n, k) = n!/(n−k)!.
- Factorial – The product of every whole number up to n, written n!. It counts full arrangements of n items and is the engine behind every counting formula.
- Repetition – Whether a chosen item returns to the pool. Without repetition each item is available at most once; with repetition items can be chosen again, as in password characters or DNA bases.
- n choose k – The spoken name for C(n, k), the binomial coefficient counting unordered selections of k items from n.
- Pascal’s triangle – The triangular array whose row n lists C(n, 0) through C(n, n). Each entry is the sum of the two above, and each row sums to a power of two.
- Binomial coefficient – The number C(n, k), named for its role as the coefficient of the a-to-the-(n−k) b-to-the-k term in the expansion of (a + b) to the n.
- Multiplicative formula – The method computing C(n, k) as a product of k fractions, avoiding the huge intermediate values that direct factorial division produces.
- Falling factorial – The product n × (n−1) × … × (n−k+1) that equals P(n, k), the count of ordered selections without repetition.
- Key space – The total count of possible keys or passwords in a scheme, a with-repetition permutation that determines brute-force resistance.
- Stars and bars – The construction turning a with-repetition combination into a without-repetition one by imagining k stars among n−1 dividers, giving C(n+k−1, k).
- Stirling’s approximation – The formula n! ≈ √(2πn)(n/e)ⁿ estimating factorials too large for exact computation, the working tool beyond this calculator’s 170 ceiling.
How to Use?
- 1
Choose what you are counting
Select Combination when order does not matter, Permutation when it does, or Factorial to count arrangements of everything at once. The swap test above the form settles the choice: if rearranging the picked items changes the outcome, use Permutation.
- 2
Enter the pool and the pick
Type the pool size n and the pick size k as whole numbers – 49 and 6 for the lottery draw, 10 and 3 for the podium. Inputs are validated as whole numbers, and k beyond n is flagged as an error in the without-repetition setting.
- 3
Set the repetition rule
Keep Without Repetition for deals, hands, and draws from a single pool. Switch to With Repetition for passwords, sequences, and scoops – anything where an item can be picked again.
- 4
Read the result
The exact count appears with its formula label, such as C(49, 6). Digit grouping keeps large counts readable – 13,983,816 rather than a scientific blob.
- 5
Compare the repetition settings
The repetition comparison output shows the same n and k under the opposite repetition rule. To see the order-matters premium too, switch the mode selector and recalculate: the same numbers run through the other formula in one click.
- 6
Check Pascal’s row
The bar chart in the results plots row n of Pascal’s triangle with your k highlighted, showing how your selection sits within the row’s full landscape of subset sizes.
- 7
Apply the one-in-X line
When outcomes are equally likely, the probability line converts your count to odds of a single specific selection – the lottery’s 1 in 13,983,816 directly.
Frequently Asked Questions
Order. A combination is a selection where the arrangement of the chosen items does not matter – a committee of 3 people is one combination no matter how you list them. A permutation is an arrangement where order is part of the outcome – gold, silver, bronze is three different permutations of the same three medallists. The two counts relate by exactly k!: choosing 3 from 10 gives 120 combinations but 720 permutations, because each set of three admits 3! = 6 orders.
Enter n as the total items, k as the chosen items, select Combination (nCr) and Without Repetition, and the calculator returns C(n, k) = n!/[k!(n−k)!] using exact integer arithmetic. The repetition comparison shows the with-repetition count alongside, and the one-in-X line converts the result to the probability of a single specific selection. For 49 choose 6 the answer is 13,983,816.
Enter n and k, select Permutation (nPr) and Without Repetition. The calculator returns P(n, k) = n!/(n−k)! exactly. The podium example: 10 runners and 3 medals give P(10, 3) = 720 ordered podium outcomes, of which each set of three medallists accounts for 6.
It means a selected item returns to the pool and can be picked again. Passwords are the clearest case: each of the 62 letters and digits stays available for every character slot, so a 5-character password space is 62 to the 5th power, not a shrinking product. For combinations, repetition changes the formula to C(n+k−1, k): choosing 3 scoops from 10 flavours with repeats gives C(12, 3) = 220 cones.
One in 13,983,816. The draw picks 6 distinct numbers from 49 where order does not matter, which is C(49, 6) = 13,983,816 equally likely tickets. Enter 49 and 6 in the Combination mode of this calculator and the one-in-X output prints the odds directly.
Because without repetition it is impossible to select more items than the pool holds – you cannot deal 7 cards from a 6-card deck. The calculator treats it as an invalid input and flags the error rather than returning a misleading zero. With repetition enabled, k can exceed n freely, since items return to the pool after each pick.
One. By convention 0! = 1, and the convention is not arbitrary: it makes every counting formula work at its boundary – C(n, 0) counts the one way to choose nothing, and the empty product of no factors is defined as 1 so that n! = n × (n−1)! continues to hold at n = 1. The factorial mode of this calculator returns 1 for 0, matching every textbook.
Because row n’s entries count the subsets of an n-item set by size. C(n, 0) counts the empty subset, C(n, 1) the singletons, through C(n, n) the full set, and each of the n items is independently either in or out – 2 to the power n subsets in total. Row 8, for example, totals 256.
About 8.07 × 10 to the 67th power – an 68-digit number that dwarfs the number of seconds since the Big Bang (about 4.3 × 10¹⁷). That is the count of distinct full-deck orderings, which is why no two well-shuffled decks in recorded history are believed to have ever matched. The factorial mode returns this value, and 170 is the largest factorial that fits standard exact arithmetic.
The same thing as this page’s Combination mode: it computes n choose k, C(n, k) = n!/[k!(n−k)!]. The r on the button is the k this page uses for chosen items. The companion nPr button computes the order-matters count P(n, k) = n!/(n−k)!. The risk with buttons is pressing the wrong one for your problem, which the swap test on this page prevents: if rearranging the items changes the outcome, you need nPr.
Divide one by the count, assuming every counted outcome is equally likely. The calculator prints this as its one-in-X line: 13,983,816 lottery tickets give a 1 in 13,983,816 jackpot chance. To express it as a percentage or compare it with other probabilities, the Percentage Calculator on this site does the conversion in one step.
Permutations with repetition. Every character slot independently takes any of the available symbols – order matters because "AB" and "BA" are different passwords, and repetition is allowed because nothing stops you from using the same letter twice. That makes a 5-character password over 62 symbols a count of 62 to the 5th power, which this calculator’s With Repetition permutation mode computes directly.
Up to 1000 in the combination and permutation modes, and 170 in the factorial mode. The caps exist where exact arithmetic ends: combinations stay exact through the standard number format’s integer range, and 170! is the largest factorial it holds. Results that would lose exactness are flagged rather than silently rounded, and genuinely huge counts need a dedicated big-number tool.
Counting is the first floor of probability. A count of equally likely outcomes divided into the total gives a probability; probabilities feed the Z-score Calculator for positions on the normal curve and the Sample Size Calculator for survey planning. This page does the counting, those pages do the statistics, and the chain runs from first principles to a national poll’s methodology note.
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