My high school daughter has a probability question that I can not solve as I am too far removed from any math course. I would greatly appreciate someone's help.

The problem reads: John is getting his ATM card activated. He must select a password containing 4 nonzero digits to be able to use this card. How many passwords are allowed if no digit may be used more than once?

I just Googled the question and you can get the answer online!

Surely you shouldn't be able to Google a homework question?

Thanks!

No, you really shouldn't be able to Google a homework problem - or at least not to just get the answer. Unfortunately, it exists out there and that seems to be the way many people are getting through school these days. Bad enough that they don't teach people to think anymore, let alone just having answers to everything plastered everywhere.

Amber - so you found an answer? Does that help your daughter to understand it?

amber - so you found an answer? Does that help your daughter to understand it?

Agreed.

I had students work in pairs on labs... when the lab tests rolled around, it was always clear who did the work and who rode the coat tails...

Fine to find the answer... I've peeked in the back of a text to check out the answer more than once to a math q... but when it comes to tests... you need to rewire the mind for the problem solving skills needed.

So... focus not on the answer, but the pattern.

Anything less will give OK homework scores (at best) and terrible test scores.

Just in case...

_ _ _ _

Let these four blanks represent the 4 digit password. The digits that can be picked are:
1, 2, 3, 4, 5, 6, 7, 8, 9

which makes 9 different numbers.

Let's pick one randomly for the first blank. How many different numbers can we pick? We can pick one of the nine different numbers. So, in the first blank, we have 9 possibilities.

(9) _ _ _

Say... we picked the number '4'

What is left? 1, 2, 3, 5, 6, 7, 8, 9, making 8 numbers left. From how many numbers can we pick one number for the second blank? We can pick one from 8 blanks. So, it becomes 8 different possibilities for the second blank.

(9) (8) _ _

With the same reasoning, the two last blanks will have 7 and 6 different possibilities respectively.

(9) (8) (7) (6)

Now, multiply these numbers together, giving:
9 x 8 x 7 x 6 = 3024

So, there are 3024 different passwords.

I hope it helped! :)