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 |
C(52,5) = 2,598,960 — that is the total number of possible 5-card poker hands from a standard deck. If order mattered (as in permutations), the number would be P(52,5) = 311,875,200 — exactly 120 times larger because each combination of 5 cards can be arranged in 5! = 120 different orders. The relationship P(n,r) = C(n,r) × r! is the bridge between the two concepts.
The practical distinction: choosing 3 people for a committee from 10 candidates is a combination problem (C(10,3) = 120) because the committee {Alice, Bob, Carol} is the same regardless of who was chosen first. Assigning those 3 people to specific roles (president, VP, secretary) is a permutation problem (P(10,3) = 720) because Alice-as-president is different from Alice-as-VP. Whenever order or position matters, use permutations. When only the selection matters, use combinations.
| Scenario | Order Matters? | Formula | Result |
|---|---|---|---|
| 3-person committee from 10 | No | C(10,3) | 120 |
| President/VP/Sec from 10 | Yes | P(10,3) | 720 |
| 5-card poker hand | No | C(52,5) | 2,598,960 |
| Lottery: 6 from 49 | No | C(49,6) | 13,983,816 |
Standard combinations assume each item can be chosen at most once. Combinations with repetition (multisets) allow the same item to be selected multiple times, using the formula C(n+r–1, r) = (n+r–1)! / (r! × (n–1)!). Choosing 3 scoops from 5 ice cream flavors (where you can repeat flavors) gives C(5+3–1, 3) = C(7,3) = 35 combinations — compared to only C(5,3) = 10 without repetition.
The stars and bars visualization explains why: imagine 3 stars (items chosen) and 4 bars (dividers between 5 flavors). The number of ways to arrange 3 stars among 4 bars is C(7,3). This same technique solves distribution problems: how many ways can 10 identical balls be placed into 4 distinct boxes? C(10+4–1, 10) = C(13,10) = 286 ways. Use the Probability Calculator to find the likelihood of specific combinations occurring in random processes.
| Scenario | Without Repetition | With Repetition | Ratio |
|---|---|---|---|
| 2 from 4 items | C(4,2) = 6 | C(5,2) = 10 | 1.67× |
| 3 from 5 items | C(5,3) = 10 | C(7,3) = 35 | 3.50× |
| 4 from 6 items | C(6,4) = 15 | C(9,4) = 126 | 8.40× |
Pascal’s triangle is an infinite triangular array where the entry at row n, position r equals C(n,r). Row 4 contains 1, 4, 6, 4, 1 — corresponding to C(4,0) through C(4,4). Each entry equals the sum of the two entries above it: C(n,r) = C(n–1,r–1) + C(n–1,r). This recursive property makes it possible to compute large combinations without evaluating factorials directly.
The binomial theorem uses Pascal’s triangle coefficients: (a+b)ⁿ = Σ C(n,k) × aⁿ⁻ᵏ × bᵏ for k = 0 to n. Expanding (x+1)⁴ gives 1x⁴ + 4x³ + 6x² + 4x + 1 — the coefficients 1, 4, 6, 4, 1 are exactly row 4 of Pascal’s triangle. Row sums equal powers of 2 (row 4: 1+4+6+4+1 = 16 = 2⁴), which is equivalent to counting all subsets of a set with n elements.
Last Updated: Mar 26, 2026
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