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27 January, 16:23

A lock requires a 3 number combination using the numbers 0 through 9, none of which may be repeated. How many outcomes are possible?

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Answers (2)
  1. 27 January, 16:45
    0
    120

    Step-by-step explanation:

    So in your question, we have

    10C3 = 10! / [3! x (10-3) !] = 10! / [3! 7!]

    = (10 x 9 x 8 x 7!) / [ (3 x 2 x 1) 7!]

    = (720) / 6 ... because the 7! on the top and bottom cancels.

    = 120 as we expected.
  2. 27 January, 20:02
    0
    120 is your answer
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