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

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.

120 is your answer