Math question, permutations/combinations wod problem...help!?

Question

Hey guys!
I've been working on this problem, but I just can't figure it out; it's probably easy, but I'm stuck on it. Here is the problem:

An alternative plan proposed a 4 digit area code and restricted the first and second digits. How many area codes would this plan have provided? I know the answer is 1600 but I can't figure out how they got that number. I've tried both nPr and nCr in every way I could think of , but I couldn't get the answer. What exactly am I supposed to do to solve the problem? Thanks!

Restrictions are that the first digit cannot be 0 or 1 and the second digit has to be either 0 or 1

Thanks I got the answer 8x2x10x10

Answer 1

Ah ok, that's what I expected.

The first digit can't be a 0 or 1, so there are 8 choices (2, 3, 4, 5, 6, 7, 8, 9).

The second digit must be a 0 or a 1, so there are 2 choices.

You can use any digit for the other 2, so there are 10 choices for each (0-9)

8x2x10x10 = 1600

No combinations or permutations needed.

PS -- Hook em Horns! Congrats on the win.

Answer 2

I don't understand the question. It sounds to me like the first digit has 10 possible combinations 0,1,2,,,9(ok maybe 9, you would not have an area code begin with 0. I need to know more about the restrictions. Otherwise 10 possibilities for each digits 4 digits = 10^4 = 10,000 or 9000 of them new.

Answer 3

What you need to use is called multiplication principle.

For the first digit, you have C(8,1) = 8 choices.
For the second digit, you have C(2,1) = 2 choices.
For the third digit, you have C(10,1) = 10 choices.
For the fourth digit, you have C(10,1) = 10 choices.

Since they are independent, by multiplication principle, you have

10x10x2x8 = 1600