Probabilty with phone numbers?

[za1128]

Okay so this is a question on my stats homework and I can't figure it out.

"Until recently, with the advent of cellular phones,modems, and pagers, the area codes in the U.S. and Canada followed a certain system. The firrst number could not be 0 or 1, the second number could only be 0 or 1, and the third could not be 0. Under this system, how many area codes are possible?"

Please please walk me through how its worked out!!!

 

[Ishuda]

To start with there are 10 numbers we can use: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9
An area code has three digits.
Since the first digit can not be a 0 or 1, we have only 8 numbers for the first digit.
Since the second digit can only be a 0 or 1, we have two numbers for the second digit.
The third digit can not be a 0 so we have 9 numbers for the third digit.
Thus there are
8 * 2 * 9 = 144 area codes under the old system.

You might like to read
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
It's a fairly short read and makes clear the difference between combinations and permutations with and without replacement which this type of problem falls under.

 

[za1128]

Thank you so much! I think I ended up overthinking it and confusing myself!

 

[elancee]

To start with there are 10 numbers we can use: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9
An area code has three digits.
Since the first digit can not be a 0 or 1, we have only 8 numbers for the first digit.
Since the second digit can only be a 0 or 1, we have two numbers for the second digit.
The third digit can not be a 0 so we have 9 numbers for the third digit.
Thus there are
8 * 2 * 9 = 144 area codes under the old system.

You might like to read
http://www.mathsisfun.com/combinatorics/combinations-permutations.html
It's a fairly short read and makes clear the difference between combinations and permutations with and without replacement which this type of problem falls under.

Thanks for posting that site!