Question

{may you please help me with this attachment}

Answer 1

what attachment?

Answer 2

thos one

Answer 3

is it 588?

Answer 4

nope i was on a mastery test lol

Answer 5

does it matter if the 3 numbers are the same? Or do all 3 numbers have to be different?

Answer 6

all 3 of dem has to be different

Answer 7

784?

Answer 8

that makes it a little more tricky, lol. It's not as simple as 14x4x14... it's a slightly more complicated combination.

Answer 9

do you have the answer? I got 325.

Answer 10

I calculated it this way; and I could be very wrong, but since no one else has given solution, might as well tell you how I went about it...

I know that if the solutions have to be unique combinations, then the answer has to be less than 455. Reason being that the total number of characters we dealing with are 15 (the numbers 0-14), and we have unique combinations of 3:

\[\frac{15!}{(15-3)!3!} = 455\]

So the answer must be less than 455.

If we look at which digit "limits" the number of unique combinations, it would be the middle digit.
Assuming this digit was zero, the total number of combinations we could have where the order didn't matter would be:

14 x 13 = 182

If the middle digit was 1, 2 or 3, this would be a little different, because we can now have a combinations such as {1,1,1} , which don't count:

13 x 12 = 156 combinations for each digit, or 3 x 156 = 468 combinations.

So our total number of combinations is 182+468 = 650.
This is a little high from our estimation... what are we forgetting? Doubles!
(ie. {12,3,6} and {6,3,12} are not unique when order doesn't matter)

so:
650/2 = 325...

Now I'm not 100% sure if this is correct... I'm not sure if I'm forgetting something, but if I am, somebody, please correct me lol.