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    20 Choose 2

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

    What is 20 choose 2?

    Answer

    20 choose 2 is
    190

    Order doesn't matter, no repeats: 190 ways to choose 2 of 20 items.

    C(20, 2) = 20! / (2! · 18!)

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

    = 190

    If order matters (permutations): P(20, 2) = 20 × 19 = 380 — each combination in 2! = 2 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.

    See Also

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

    n choose 2 for other n

    n choose 2Answer
    6 choose 215
    7 choose 221
    8 choose 228
    9 choose 236
    10 choose 245
    11 choose 255
    12 choose 266
    13 choose 278
    14 choose 291
    15 choose 2105
    16 choose 2120
    17 choose 2136
    18 choose 2153
    19 choose 2171
    20 choose 2190
    Recalculate with different values ↑

    FAQ

    What is 20 choose 2?

    20 choose 2 is 190

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

    Every way to choose 2 of 20 items leaves exactly 18 behind, so choosing 2 and choosing which 18 to leave out are the same count: 20 choose 2 = 20 choose 18 = 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, 2) = 380. Each of the 190 combinations can be put in order in 2! = 2 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.

    See Also