Case 1
- Total items (n)
- 10
- Items chosen (r)
- 3
120
Open with these valuesCombinations Calculator
120combinations
Result: 120 combinationsA combination counts selections, not orders: picking Alice, Bob and Carol is the same as picking Carol, Alice and Bob. C(10, 3) = 120 ways to choose three people from ten. The same three in a different order would be a permutation — there are 720 of those.
The numbers at a glance
Held fixed: Total items (n) 10.
| Items chosen (r) | Result |
|---|---|
| 0 | 1 |
| 1 | 10 |
| 2 | 45 |
| 3Your value | 120 |
| 4 | 210 |
| 5 | 252 |
| 6 | 210 |
Worked examples
Case 2
- Total items (n)
- 49
- Items chosen (r)
- 6
13,983,816
Open with these valuesCase 3
- Total items (n)
- 52
- Items chosen (r)
- 5
2,598,960
Open with these valuesHow it's calculated
C(n, r) = n! ÷ (r! × (n − r)!)
- StepEnter the size of the whole set as n — ten students, 49 lottery numbers, 52 cards.
- StepEnter how many you take as r; the order you take them in is ignored.
- ResultRead the number of distinct groups. Asking for more than exists gives 0.
What this number means
A combination counts groups. What is interesting is rarely which three of ten people end up on a committee but how many such committees exist at all, because that number sets the odds on any one of them and it grows faster than intuition allows. The formula reads most easily backwards, as the correction of an easier count. Filling three ranked places from ten people is a plain product: ten candidates for the first, nine for the second, eight for the third, so 10 × 9 × 8 = 720. But a committee has no ranks, and every group of three turns up in that product once for each order its members can stand in — 3! = 6 times. Dividing removes exactly that duplication: 720 ÷ 6 = 120. That is where the r! in the denominator comes from, and because every group was over-counted by the same factor of 6, the division always comes out whole. What the count does not do is tell the groups apart: all 13,983,816 six-from-49 tickets are one single number here, and that number says nothing about how any of them came about. The practical ceiling is arithmetic — above n = 55 a standard number can no longer write these coefficients out digit for digit, and C(56, 23) prints one too high.
Order ignored, and nothing is drawn twice
Each item is used up when you take it, and the sequence of the picks is never counted. A six-from-49 lottery ticket is the standard case: 13,983,816 tickets, and no number appears twice on one.
C(n, r) is P(n, r) divided by r factorial
Every selection of r items can be arranged in r factorial orders, so the combination count is always the smaller of the two. Three from ten gives 120 combinations against 720 permutations — exactly the factor 3 factorial, which is 6.
More than exists gives 0, not an error
With r greater than n the answer is 0, because no such selection can be made. That zero is the correct count, not a refusal to answer.
Commonly misread
C(10, 3) and C(10, 7) must be different numbers.
Choosing 3 to keep also decides which 7 stay behind, so both are 120. The calculator evaluates whichever of the two is smaller.
The 120 committees of three from ten already fix who chairs.
C(10, 3) counts groups only. Naming a chair, a deputy and an ordinary member ranks the three and gives 720 instead.
C(5, 0) is 0, because nothing was chosen.
It is 1: there is exactly one way to choose nothing, and exactly one way to take everything. Only r greater than n gives 0.
Reference table
| n, r | Reads as | C(n, r) |
|---|---|---|
| 0, 0 | choose nothing from nothing | 1 |
| 5, 0 | the empty selection | 1 |
| 5, 5 | take the whole set | 1 |
| 10, 3 | a committee of three from ten | 120 |
| 20, 10 | half of twenty | 184756 |
| 49, 6 | a six-from-49 lottery ticket | 13983816 |
| 52, 5 | a five-card poker hand | 2598960 |
Questions
What is a combination?
A combination is a selection in which the order does not matter, so Alice, Bob, Carol is the same choice as Carol, Alice, Bob. The count is written C(n, r), spoken n choose r, and equals n!/(r!(n−r)!). Choosing 3 from 10 gives 120.
How are combinations different from permutations?
Permutations count arrangements where order matters; combinations count selections where it does not. Every combination of r items can be arranged in r factorial different orders, so there are never fewer permutations than combinations. Choosing 3 from 10 gives 120 combinations but 720 permutations.
Why does C(n, r) equal C(n, n − r)?
Choosing r items to keep also decides which n − r items to leave behind, so both questions have the same answer. That is why C(10, 3) and C(10, 7) are both 120. This calculator evaluates whichever of the two is smaller.
What are C(n, 0) and C(n, n)?
Both are 1: there is exactly one way to choose nothing, and exactly one way to take everything. If you ask for more items than exist, with r greater than n, the answer is 0, because no such selection can be made.
Where are combinations used?
Anywhere a group is picked without ranking it: lottery tickets (6 from 49 gives 13,983,816), poker hands (5 from 52 gives 2,598,960), committees, pizza toppings. The same numbers are the binomial coefficients of the binomial theorem.
Sources and last check
Information, not professional advice.