Question

I have a GRE question:
How many different 3-digit numbers can be formed, using the digits from 0 to 9 inclusive, if no digit is used more than once?
A 504
B 648
C 729
D 900
E 1,000

Answer 1

Do you think it would be easier to count the number of 3 digit numbers with repeats, or the number of 3 digit numbers with no repeats?

Answer 2

This is actually just a simple permutation task, isn't it?

Answer 3

Including repeats the answer is 1,000
but the question says "no repeats"
Something like 10!/7! is what I figure the answer should be, but given the 5 solutions, only option C comes close.

Answer 4

I see, you can't use 0 as a leading digit, that is what makes it complicated.

I would start with 10 P 7, then try to remove the leading 0 numbers.

Answer 5

We know with 10 P 7, we won't have anything like 001, But we will have something like 012

Answer 6

Okay I've been saying 10 P 7, I mean 10 P 3

Answer 7

Assuming the first digit is 0 though, the remaining qualifiers are 9 P 2

Answer 8

So then you have 10 P 3 - 9 P 2 = 10!/7! - 9!/7! =10 * 9!/7! - 9!/7! = 9*9!/7!

Answer 9

@Matt.Mawson I end up getting an answer that is listed.