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    20 Choose 18

    Combination Calculator
    Total items (n)
    Items chosen (k)

    What is 20 choose 18?

    Answer

    20 choose 18 is
    190

    Order doesn't matter, no repeats: 190 ways to choose 18 of 20 items.

    Choosing 18 of 20 is the same as leaving 2 out: C(20, 18) = C(20, 2).

    C(20, 18) = 20! / (18! · 2!)

    = C(20, 2)

    = (20 × 19) / (2 × 1)
    = 380 / 2

    = 190

    If order matters (permutations): P(20, 18) = 20 × 19 × 18 × … × 3 = 1,216,451,004,088,320,000 — each combination in 18! = 6,402,373,705,728,000 orders.

    Row 20 of Pascal's triangle
    k20 choose k
    01
    120
    2190
    31,140
    44,845
    515,504
    638,760
    777,520
    8125,970
    9167,960
    10184,756
    11167,960
    12125,970
    1377,520
    1438,760
    1515,504
    164,845
    171,140
    18190
    1920
    201

    Row sum: 220 = 1,048,576 — all subsets of 20 items.

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    https://calculat.io/en/number/combinations/20--18Copy
    <a href="https://calculat.io/en/number/combinations/20--18">20 Choose 18: Combinations C(20, 18) - Calculatio</a>Copy
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    20 Choose 18: Combinations C(20, 18). 20 choose 18 = 190: the number of ways to choose 18 of 20 items when order doesn't matter and nothing repeats. C(20, 18) = (20 × 19) / (2 × 1).
    https://calculat.io/en/number/combinations/20--18/generated.jpgCopy

    n choose 18 for other n

    n choose 18Answer
    18 choose 181
    19 choose 1819
    20 choose 18190
    Recalculate with different values ↑

    FAQ

    What is 20 choose 18?

    20 choose 18 is 190

    Why is 20 choose 18 the same as 20 choose 2?

    Every way to choose 18 of 20 items leaves exactly 2 behind, so choosing 18 and choosing which 2 to leave out are the same count: 20 choose 18 = 20 choose 2 = 190. In the formula n! / (k! (n − k)!) the two factorials below just swap places.

    How many ways are there if order matters?

    P(20, 18) = 1,216,451,004,088,320,000. Each of the 190 combinations can be put in order in 18! = 6,402,373,705,728,000 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.

    See Also