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    Combination Calculator

    How many ways to choose k of n items when order doesn’t matter, with the formula worked out

    Combination Calculator

    Total items (n)
    Items chosen (k)

    How to Count Combinations: n Choose k

    C(n, k) = n! / (k! (n − k)!): multiply k numbers counting down from n, then divide by k!.

    Example: 4 choose 2

    C(4, 2) = 4! / (2! · 2!)

    = (4 × 3) / (2 × 1)
    = 12 / 2

    = 6

    Common mistakes

    • k larger than n gives 0, not an error: 1 choose 3 = 0, because you can't choose 3 different items from 1. Usually the numbers were swapped — 3 choose 1 = 3.
    • 0! = 1, so choosing none and choosing all are one way each: 5 choose 0 = 5 choose 5 = 1.
    • If order matters, count permutations P(n, k) = n! / (n − k)! instead: 4 choose 2 = 6, but P(4, 2) = 12 — each pair counted in 2 orders.
    • If items may repeat (3 scoops from 4 flavours, two of a kind allowed), use C(n + k − 1, k): C(6, 3) = 20, not 4 choose 3 = 4.
    • nCr, ⁿCₖ, C(n, k) and the binomial coefficient “n over k” are the same number: 4C2 = C(4, 2) = 6.

    See Also

    Last Results

    18 Choose 4: Combinations C(18, 4)
    4 Choose 0: Combinations C(4, 0)
    18 Choose 9: Combinations C(18, 9)
    17 Choose 17: Combinations C(17, 17)
    12 Choose 10: Combinations C(12, 10)
    16 Choose 16: Combinations C(16, 16)
    16 Choose 15: Combinations C(16, 15)
    18 Choose 16: Combinations C(18, 16)
    15 Choose 7: Combinations C(15, 7)
    13 Choose 7: Combinations C(13, 7)
    10 Choose 4: Combinations C(10, 4)
    1 Choose 3: Combinations C(1, 3)
    13 Choose 2: Combinations C(13, 2)
    14 Choose 7: Combinations C(14, 7)
    15 Choose 13: Combinations C(15, 13)
    16 Choose 6: Combinations C(16, 6)
    17 Choose 15: Combinations C(17, 15)
    13 Choose 3: Combinations C(13, 3)
    15 Choose 15: Combinations C(15, 15)
    11 Choose 3: Combinations C(11, 3)
    17 Choose 16: Combinations C(17, 16)
    15 Choose 5: Combinations C(15, 5)
    4 Choose 1: Combinations C(4, 1)
    15 Choose 2: Combinations C(15, 2)
    15 Choose 4: Combinations C(15, 4)
    19 Choose 4: Combinations C(19, 4)
    16 Choose 12: Combinations C(16, 12)
    18 Choose 5: Combinations C(18, 5)
    11 Choose 10: Combinations C(11, 10)
    10 Choose 10: Combinations C(10, 10)
    12 Choose 8: Combinations C(12, 8)
    19 Choose 17: Combinations C(19, 17)
    15 Choose 6: Combinations C(15, 6)
    16 Choose 1: Combinations C(16, 1)
    17 Choose 3: Combinations C(17, 3)
    18 Choose 11: Combinations C(18, 11)
    1 Choose 2: Combinations C(1, 2)
    2 Choose 10: Combinations C(2, 10)
    2 Choose 0: Combinations C(2, 0)
    14 Choose 14: Combinations C(14, 14)
    14 Choose 12: Combinations C(14, 12)
    20 Choose 8: Combinations C(20, 8)
    20 Choose 10: Combinations C(20, 10)
    16 Choose 8: Combinations C(16, 8)
    3 Choose 2: Combinations C(3, 2)
    20 Choose 18: Combinations C(20, 18)
    19 Choose 12: Combinations C(19, 12)
    19 Choose 0: Combinations C(19, 0)
    15 Choose 8: Combinations C(15, 8)
    11 Choose 11: Combinations C(11, 11)
    18 Choose 3: Combinations C(18, 3)
    14 Choose 2: Combinations C(14, 2)
    20 Choose 15: Combinations C(20, 15)
    19 Choose 3: Combinations C(19, 3)
    20 Choose 13: Combinations C(20, 13)
    14 Choose 10: Combinations C(14, 10)

    FAQ

    What is the combination formula?

    C(n, k) = n! / (k! (n − k)!). The factorials cancel down to a short product: multiply k numbers counting down from n and divide by k!. For 5 choose 2 that is (5 × 4) / (2 × 1) = 10.

    What is the difference between combinations and permutations?

    Combinations ignore order — {A, B} and {B, A} are one choice; permutations count every order separately. So P(n, k) = C(n, k) × k!: from 5 items there are 10 pairs but 20 ordered pairs.

    Why is C(n, k) equal to C(n, n − k)?

    Choosing k items is the same as choosing the n − k you leave out. So 10 choose 8 = 10 choose 2 = 45, and it is quicker to compute through the smaller k.