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    10 Choose 6

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

    What is 10 choose 6?

    Answer

    10 choose 6 is
    210

    Order doesn't matter, no repeats: 210 ways to choose 6 of 10 items.

    Choosing 6 of 10 is the same as leaving 4 out: C(10, 6) = C(10, 4).

    C(10, 6) = 10! / (6! · 4!)

    = C(10, 4)

    = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1)
    = 5,040 / 24

    = 210

    If order matters (permutations): P(10, 6) = 10 × 9 × 8 × 7 × 6 × 5 = 151,200 — each combination in 6! = 720 orders.

    Row 10 of Pascal's triangle
    k10 choose k
    01
    110
    245
    3120
    4210
    5252
    6210
    7120
    845
    910
    101

    Row sum: 210 = 1,024 — all subsets of 10 items.

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    https://calculat.io/en/number/combinations/10--6Copy
    <a href="https://calculat.io/en/number/combinations/10--6">10 Choose 6: Combinations C(10, 6) - Calculatio</a>Copy
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    10 Choose 6: Combinations C(10, 6). 10 choose 6 = 210: the number of ways to choose 6 of 10 items when order doesn't matter and nothing repeats. C(10, 6) = (10 × 9 × 8 × 7) / (4 × 3 × 2 × 1).
    https://calculat.io/en/number/combinations/10--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 10 choose 6?

    10 choose 6 is 210

    Why is 10 choose 6 the same as 10 choose 4?

    Every way to choose 6 of 10 items leaves exactly 4 behind, so choosing 6 and choosing which 4 to leave out are the same count: 10 choose 6 = 10 choose 4 = 210. In the formula n! / (k! (n − k)!) the two factorials below just swap places.

    How many ways are there if order matters?

    P(10, 6) = 151,200. Each of the 210 combinations can be put in order in 6! = 720 ways, so permutations = combinations × k!: P(n, k) = n! / (n − k)!.

    See Also