MHB More statistics: counting problem

crystal1
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"For years, telephone area codes in the U.S. and Canada consisted of a sequence of three digits. The first digit was an integer between 2 and 9, the second digit was either 0 or 1, and the third digit was any integer from 1 to 9.
(1) How many area codes were possible?
(2) How many area codes starting with a 4 were possible?

Again, there are no further instructions and I do not know which formula to use for this problem.
 
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This is an application of the fundamental counting principle, much like your previous problem.

For part 1), can you state how many choices there are for each of the digits, and then find the product of these numbers of choices?
 
These are my choices..
2,3,4,5,6,7,8,9
0,1
1,2,3,4,5,6,7,8,9

However, I am unsure if multiplying them all together like the last problem would work? Or if I multiply the number of choices like 8*2*9?
 
crystal said:
These are my choices..
2,3,4,5,6,7,8,9
0,1
1,2,3,4,5,6,7,8,9

However, I am unsure if multiplying them all together like the last problem would work? Or if I multiply the number of choices like 8*2*9?

Your counts and application of the fundamental counting principle are exactly right. (Yes)

How would you now do part 2)?
 
6*6?
 
crystal said:
6*6?

For part 2), the first digit must be a 4, so we only have 1 choice there, and the remaining two digits have the same choices as they did for part 1). So, how many possible area codes do we have?
 
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