rositsaivanova1 rositsaivanova1

24-04-2019
Mathematics

contestada

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

Respuesta :

sneakerplugaj sneakerplugaj

24-04-2019

Answer:

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.

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Аноним Аноним

24-04-2019

Answer:

120 is your answer

Answer Link

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