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    15 Choose 5

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

    What is 15 choose 5?

    Answer

    15 choose 5 is
    3,003

    Order doesn't matter, no repeats: 3,003 ways to choose 5 of 15 items.

    C(15, 5) = 15! / (5! · 10!)

    = (15 × 14 × 13 × 12 × 11) / (5 × 4 × 3 × 2 × 1)
    = 360,360 / 120

    = 3,003

    If order matters (permutations): P(15, 5) = 15 × 14 × 13 × 12 × 11 = 360,360 — each combination in 5! = 120 orders.

    Row 15 of Pascal's triangle
    k15 choose k
    01
    115
    2105
    3455
    41,365
    53,003
    65,005
    76,435
    86,435
    95,005
    103,003
    111,365
    12455
    13105
    1415
    151

    Row sum: 215 = 32,768 — all subsets of 15 items.

    See Also

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    https://calculat.io/en/number/combinations/15--5Copy
    <a href="https://calculat.io/en/number/combinations/15--5">15 Choose 5: Combinations C(15, 5) - Calculatio</a>Copy
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    15 Choose 5: Combinations C(15, 5). 15 choose 5 = 3,003: the number of ways to choose 5 of 15 items when order doesn't matter and nothing repeats.
    https://calculat.io/en/number/combinations/15--5/generated.jpgCopy

    n choose 5 for other n

    n choose 5Answer
    6 choose 56
    7 choose 521
    8 choose 556
    9 choose 5126
    10 choose 5252
    11 choose 5462
    12 choose 5792
    13 choose 51,287
    14 choose 52,002
    15 choose 53,003
    16 choose 54,368
    17 choose 56,188
    18 choose 58,568
    19 choose 511,628
    20 choose 515,504
    Recalculate with different values ↑

    FAQ

    What is 15 choose 5?

    15 choose 5 is 3,003

    Why is 15 choose 5 the same as 15 choose 10?

    Every way to choose 5 of 15 items leaves exactly 10 behind, so choosing 5 and choosing which 10 to leave out are the same count: 15 choose 5 = 15 choose 10 = 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(15, 5) = 360,360. Each of the 3,003 combinations can be put in order in 5! = 120 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.

    See Also