C(n, r)
120
n
10
r
3
P(n,r)
720
Each item can only be selected once. Formula: n! / (r!(n-r)!)
120
220
C(10, 3) = 10! / (3! × (10 - 3)!)
= 10! / (3! × 7!)
= 3,628,800 / (6 × 5,040)
= 120
Highlighted value at position r=3 is C(10, 3) = 120
C(n, r) = n! / (r! × (n - r)!)Count the number of ways to choose r items from n distinct items where order does not matter.
Where:
n= Total number of items to choose fromr= Number of items being chosenn!= n factorial = n × (n-1) × ... × 1C_rep(n, r) = (n + r - 1)! / (r! × (n - 1)!)Count the number of ways to choose r items from n types when items can be reused (multiset selections).
Where:
n= Number of item types availabler= Number of items being chosenC(n, r) = P(n, r) / r!Combinations equal permutations divided by r!, because order does not matter in combinations.
Where:
P(n,r)= Number of permutations (ordered arrangements)r!= Number of ways to arrange r itemsInputs
Result
C(10,3) = 10!/(3!×7!) = (10×9×8)/(3×2×1) = 720/6 = 120. There are 120 ways to form a 3-person committee from 10 candidates.
Inputs
Result
C(52,5) = 52!/(5!×47!) = 2,598,960. There are nearly 2.6 million possible 5-card poker hands from a standard deck.
Inputs
Result
C(6,6) = 6!/(6!×0!) = 1. There is only one way to choose all items: take them all. However, P(6,6) = 720 because there are 720 orderings.
A combination counts the number of ways to select items where order does not matter. Choosing {A, B, C} is the same combination as {C, B, A}. In permutations, order matters, so ABC and CBA are counted separately. The formula is C(n,r) = n! / (r!(n-r)!).
| Scenario | Combination C(n,r) | Permutation P(n,r) |
|---|---|---|
| 3 from 10 | 120 | 720 |
| 5 from 52 (poker) | 2,598,960 | 311,875,200 |
| 2 from 8 | 28 | 56 |
| 6 from 49 (lottery) | 13,983,816 | 10,068,347,520 |
Combinations with repetition (multisets) use the formula C(n+r-1, r) = (n+r-1)! / (r!(n-1)!). This counts selections where the same item can be chosen multiple times. For example, choosing 3 scoops from 5 ice cream flavors: C(5+3-1, 3) = C(7,3) = 35 combinations.
| Scenario | Without Repetition | With Repetition |
|---|---|---|
| 3 from 5 items | C(5,3) = 10 | C(7,3) = 35 |
| 2 from 4 items | C(4,2) = 6 | C(5,2) = 10 |
| 4 from 6 items | C(6,4) = 15 | C(9,4) = 126 |
Pascal's triangle is a triangular array where each entry is C(n,r), the number of combinations of r items from n. Each number equals the sum of the two numbers above it. Row n contains all values of C(n,0) through C(n,n). It is used in probability, algebra, and binomial expansions.
"n choose r", written as C(n,r) or ⁷Cr, is the binomial coefficient. It represents the number of ways to choose r items from a set of n distinct items, without regard to order. The formula is C(n,r) = n! / (r! × (n-r)!).
| Expression | Expansion | Result |
|---|---|---|
| C(5,2) | 5!/(2!×3!) | 10 |
| C(10,3) | 10!/(3!×7!) | 120 |
| C(20,5) | 20!/(5!×15!) | 15,504 |
| C(52,5) | 52!/(5!×47!) | 2,598,960 |
Combinations are a core concept in combinatorics, counting the number of ways to select items from a larger set when the order of selection does not matter. The combination formula C(n,r) = n!/(r!(n-r)!) gives the binomial coefficient, which appears throughout mathematics, from probability theory to polynomial expansions.
Pascal's triangle provides a visual representation of all binomial coefficients. Each row n contains the values C(n,0) through C(n,n), and the recursive property C(n,r) = C(n-1,r-1) + C(n-1,r) means each entry is the sum of the two entries above it. This elegant pattern connects to the binomial theorem (a+b)^n.
Combinations with repetition (multichoose) extend the concept to situations where items can be selected multiple times. The formula C(n+r-1, r) uses the stars and bars method and is essential in problems like distributing identical objects into distinct groups or counting multisets.
Last Updated: Mar 9, 2026
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