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    14 Choose 6

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

    What is 14 choose 6?

    Answer

    14 choose 6 is
    3,003

    Order doesn't matter, no repeats: 3,003 ways to choose 6 of 14 items.

    C(14, 6) = 14! / (6! · 8!)

    = (14 × 13 × 12 × 11 × 10 × 9) / (6 × 5 × 4 × 3 × 2 × 1)
    = 2,162,160 / 720

    = 3,003

    If order matters (permutations): P(14, 6) = 14 × 13 × 12 × 11 × 10 × 9 = 2,162,160 — each combination in 6! = 720 orders.

    Row 14 of Pascal's triangle
    k14 choose k
    01
    114
    291
    3364
    41,001
    52,002
    63,003
    73,432
    83,003
    92,002
    101,001
    11364
    1291
    1314
    141

    Row sum: 214 = 16,384 — all subsets of 14 items.

    See Also

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    https://calculat.io/en/number/combinations/14--6Copy
    <a href="https://calculat.io/en/number/combinations/14--6">14 Choose 6: Combinations C(14, 6) - Calculatio</a>Copy
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    14 Choose 6: Combinations C(14, 6). 14 choose 6 = 3,003: the number of ways to choose 6 of 14 items when order doesn't matter and nothing repeats.
    https://calculat.io/en/number/combinations/14--6/generated.jpgCopy

    n choose 6 for other n

    n choose 6Answer
    6 choose 61
    7 choose 67
    8 choose 628
    9 choose 684
    10 choose 6210
    11 choose 6462
    12 choose 6924
    13 choose 61,716
    14 choose 63,003
    15 choose 65,005
    16 choose 68,008
    17 choose 612,376
    18 choose 618,564
    19 choose 627,132
    20 choose 638,760
    Recalculate with different values ↑

    FAQ

    What is 14 choose 6?

    14 choose 6 is 3,003

    Why is 14 choose 6 the same as 14 choose 8?

    Every way to choose 6 of 14 items leaves exactly 8 behind, so choosing 6 and choosing which 8 to leave out are the same count: 14 choose 6 = 14 choose 8 = 3,003. In the formula n! / (k! (n − k)!) the two factorials below just swap places.

    How many ways are there if order matters?

    P(14, 6) = 2,162,160. Each of the 3,003 combinations can be put in order in 6! = 720 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.

    See Also