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    20 Choose 3

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

    What is 20 choose 3?

    Answer

    20 choose 3 is
    1,140

    Order doesn't matter, no repeats: 1,140 ways to choose 3 of 20 items.

    C(20, 3) = 20! / (3! · 17!)

    = (20 × 19 × 18) / (3 × 2 × 1)
    = 6,840 / 6

    = 1,140

    If order matters (permutations): P(20, 3) = 20 × 19 × 18 = 6,840 — each combination in 3! = 6 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--3Copy
    <a href="https://calculat.io/en/number/combinations/20--3">20 Choose 3: Combinations C(20, 3) - Calculatio</a>Copy
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    20 Choose 3: Combinations C(20, 3). 20 choose 3 = 1,140: the number of ways to choose 3 of 20 items when order doesn't matter and nothing repeats. C(20, 3) = (20 × 19 × 18) / (3 × 2 × 1).
    https://calculat.io/en/number/combinations/20--3/generated.jpgCopy

    n choose 3 for other n

    n choose 3Answer
    9 choose 384
    10 choose 3120
    11 choose 3165
    12 choose 3220
    13 choose 3286
    14 choose 3364
    15 choose 3455
    16 choose 3560
    17 choose 3680
    18 choose 3816
    19 choose 3969
    20 choose 31,140
    25 choose 32,300
    40 choose 39,880
    100 choose 3161,700
    Recalculate with different values ↑

    FAQ

    What is 20 choose 3?

    20 choose 3 is 1,140

    Why is 20 choose 3 the same as 20 choose 17?

    Every way to choose 3 of 20 items leaves exactly 17 behind, so choosing 3 and choosing which 17 to leave out are the same count: 20 choose 3 = 20 choose 17 = 1,140. In the formula n! / (k! (n − k)!) the two factorials below just swap places.

    How many ways are there if order matters?

    P(20, 3) = 6,840. Each of the 1,140 combinations can be put in order in 3! = 6 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.

    See Also