How many 3-digit numbers can be made by using the numbers 1 2 3 4 5 6 7 9?

Answer 1

You have 8 numbers, and you are choosing 3. This is written as 8c3 and is computed like this:

[8!] / [ (3!) (8-3!)] = 8! / (3! * 5!) = (8*7*6*5*4*3*2*1) / (3*2*1 * 5*4*3*2*1)

Canceling out, you get (8*7*6)/(3*2*1) = 8*7 = 56 ■