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    17 Choose 3

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

    What is 17 choose 3?

    Answer

    17 choose 3 is
    680

    Order doesn't matter, no repeats: 680 ways to choose 3 of 17 items.

    C(17, 3) = 17! / (3! · 14!)

    = (17 × 16 × 15) / (3 × 2 × 1)
    = 4,080 / 6

    = 680

    If order matters (permutations): P(17, 3) = 17 × 16 × 15 = 4,080 — each combination in 3! = 6 orders.

    Row 17 of Pascal's triangle
    k17 choose k
    01
    117
    2136
    3680
    42,380
    56,188
    612,376
    719,448
    824,310
    924,310
    1019,448
    1112,376
    126,188
    132,380
    14680
    15136
    1617
    171

    Row sum: 217 = 131,072 — all subsets of 17 items.

    See Also

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    https://calculat.io/en/number/combinations/17--3Copy
    <a href="https://calculat.io/en/number/combinations/17--3">17 Choose 3: Combinations C(17, 3) - Calculatio</a>Copy
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    17 Choose 3: Combinations C(17, 3). 17 choose 3 = 680: the number of ways to choose 3 of 17 items when order doesn't matter and nothing repeats. C(17, 3) = (17 × 16 × 15) / (3 × 2 × 1).
    https://calculat.io/en/number/combinations/17--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 17 choose 3?

    17 choose 3 is 680

    Why is 17 choose 3 the same as 17 choose 14?

    Every way to choose 3 of 17 items leaves exactly 14 behind, so choosing 3 and choosing which 14 to leave out are the same count: 17 choose 3 = 17 choose 14 = 680. In the formula n! / (k! (n − k)!) the two factorials below just swap places.

    How many ways are there if order matters?

    P(17, 3) = 4,080. Each of the 680 combinations can be put in order in 3! = 6 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.

    See Also