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    14 Choose 4

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

    What is 14 choose 4?

    Answer

    14 choose 4 is
    1,001

    Order doesn't matter, no repeats: 1,001 ways to choose 4 of 14 items.

    C(14, 4) = 14! / (4! · 10!)

    = (14 × 13 × 12 × 11) / (4 × 3 × 2 × 1)
    = 24,024 / 24

    = 1,001

    If order matters (permutations): P(14, 4) = 14 × 13 × 12 × 11 = 24,024 — each combination in 4! = 24 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--4Copy
    <a href="https://calculat.io/en/number/combinations/14--4">14 Choose 4: Combinations C(14, 4) - Calculatio</a>Copy
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    14 Choose 4: Combinations C(14, 4). 14 choose 4 = 1,001: the number of ways to choose 4 of 14 items when order doesn't matter and nothing repeats. C(14, 4) = (14 × 13 × 12 × 11) / (4 × 3 × 2 × 1).
    https://calculat.io/en/number/combinations/14--4/generated.jpgCopy

    n choose 4 for other n

    n choose 4Answer
    7 choose 435
    8 choose 470
    9 choose 4126
    10 choose 4210
    11 choose 4330
    12 choose 4495
    13 choose 4715
    14 choose 41,001
    15 choose 41,365
    16 choose 41,820
    17 choose 42,380
    18 choose 43,060
    19 choose 43,876
    20 choose 44,845
    24 choose 410,626
    Recalculate with different values ↑

    FAQ

    What is 14 choose 4?

    14 choose 4 is 1,001

    Why is 14 choose 4 the same as 14 choose 10?

    Every way to choose 4 of 14 items leaves exactly 10 behind, so choosing 4 and choosing which 10 to leave out are the same count: 14 choose 4 = 14 choose 10 = 1,001. In the formula n! / (k! (n − k)!) the two factorials below just swap places.

    How many ways are there if order matters?

    P(14, 4) = 24,024. Each of the 1,001 combinations can be put in order in 4! = 24 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.

    See Also