About the Combination Calculator
This combination calculator answers “how many different groups can I make?” Enter the total number of items n and how many you choose r, and it returns nCr — the number of unordered selections — along with the matching permutation count nPr so you can see how much order matters.
Use it for lottery odds, poker hands, picking a committee or team, choosing pizza toppings, or any probability and statistics homework involving “n choose k”. Switch on repetition when the same item can be picked more than once (like scoops of ice cream from a set of flavours), which uses the multiset formula C(n + r − 1, r).
The chart shows the full row of Pascal’s triangle for your n, so you can see how the count peaks when you choose about half of the items. Very large results are shown in scientific notation because they exceed exact double-precision integers.
With the default inputs, the number of combinations is 120. Change any value above to recalculate instantly.
How to use the combination calculator
- 1Enter the total number of items n.
- 2Enter how many items r you pick for each group.
- 3Turn on repetition if an item can be picked more than once.
- 4Read the number of combinations and compare with the permutation count.
- 5Use the “1 in N” figure for lottery or card-hand odds.
Formula and method
A combination counts selections where order does not matter. The permutation count n! ÷ (n − r)! counts ordered arrangements; every unordered group of r items appears r! times among them, so dividing by r! gives nCr.
When items may repeat, the “stars and bars” argument turns the problem into choosing r positions among n + r − 1 slots, giving C(n + r − 1, r). The calculator multiplies and divides step by step instead of computing huge factorials, so it stays accurate for large n.
- n
- Total number of distinct items to choose from
- r
- Number of items in each selection
- !
- Factorial: n! = n × (n − 1) × … × 1, with 0! = 1
Worked examples
Choose 3 people from 10
There are 10 × 9 × 8 = 720 ordered ways to pick 3 people, but each trio appears 3! = 6 times, so there are 720 ÷ 6 = 120 distinct groups.
Five-card poker hands
C(52, 5) = 52! ÷ (5! × 47!) = 2,598,960 possible poker hands, so any one specific hand has a 1 in 2,598,960 chance of being dealt.
Lottery: pick 6 numbers from 49
A 6/49 lottery has C(49, 6) = 13,983,816 possible tickets, so a single ticket has roughly a 1 in 14 million chance of matching all six numbers.
3 scoops from 5 flavours, repeats allowed
With repetition the count is C(5 + 3 − 1, 3) = C(7, 3) = 35 cups, compared with only C(5, 3) = 10 if every scoop had to be a different flavour.
Frequently asked questions
What is the difference between a combination and a permutation?+
In a combination the order does not matter (a team of Ann, Ben and Cy is one group). In a permutation it does (gold, silver, bronze for Ann, Ben, Cy is different from Cy, Ben, Ann). nPr is always r! times larger than nCr.
What does n choose k mean?+
“n choose k” is another name for C(n, k), the number of ways to pick k items from n without regard to order. It is also entry k in row n of Pascal’s triangle (counting rows and entries from 0) and the coefficient in the binomial expansion.
What is C(n, 0) and C(n, n)?+
Both equal 1. There is exactly one way to choose nothing and exactly one way to choose everything, which is why 0! is defined as 1.
How do I calculate combinations with repetition?+
Use C(n + r − 1, r). For example, choosing 3 donuts from 4 flavours where flavours may repeat gives C(6, 3) = 20 possible boxes.
Why is C(n, r) equal to C(n, n − r)?+
Choosing which r items to take is the same as choosing which n − r items to leave behind, so both counts are identical. That is why each row of Pascal’s triangle is symmetric.