SOLUTION: the word problem is " Mosley Hills has reserved the digits 444, 445, and 446 as the prefixes for their phone number. How many different phone numbers are possible?

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Question 116189: the word problem is " Mosley Hills has reserved the digits 444, 445, and 446 as the prefixes for their phone number. How many different phone numbers are possible?
Found 2 solutions by heidi, Edwin McCravy:
Answer by heidi(17)   (Show Source): You can put this solution on YOUR website!
444, 445, 446
444,446, 445
446, 444, 445
446, 445,444
445,444,446
445, 446, 444
(6 different numbers )

Answer by Edwin McCravy(20055)   (Show Source): You can put this solution on YOUR website!

HEIDI MEANT WELL BUT SHE
MISINTERPRETED THE PROBLEM.
SHE THOUGHT IT WAS
"How many ways can you arrange
those three prefixes in a row?".
`
HERE IS THE CORRECT SOLUTION:

the word problem is " Mosley Hills has reserved the digits 444, 445, and 446
as the prefixes for their phone number. How many different phone numbers are
possible?

Here are three methods.  The way to do it depends on what you are
studying.

Method 1. 

The number of integers from integer M to integer N, inclusive of both, where
M < N is N - M + 1. (You have to add the 1 to N - M because when you subtract
M from N, you are discounting all the numbers up through M, so you have to 
add 1 to recount the M.)

The smallest phone number possible is 4440000.  The largest phone number
possible is 4469999.  So the number of phone numbers is

4469999 - 4440000 + 1 = 30000

-------------------------------

Method 2.

Pick a phone number at random to look at as you think:

445-7338

There is 1 choice for the 1st digit, namely 4.
There is 1 choice for the 2nd digit, namely 4.
There are 3 choices for the 3rd digit, namely 4,5, or 6.
There are 10 choices for the 4th digit, namely 0,1,2,3,4,5,6,7,8, or 9
There are 10 choices for the 5th digit, namely 0,1,2,3,4,5,6,7,8, or 9
There are 10 choices for the 6th digit, namely 0,1,2,3,4,5,6,7,8, or 9
There are 10 choices for the 7th digit, namely 0,1,2,3,4,5,6,7,8, or 9

So the number of phone numbers is 1×1×3×10×10×10×10 = 30,000

------------------------------------

Method 3.

The phone numbers:

4440000, 4440001, ..., 4469999

form an arithmetic series with a1 = 4440000, an = 4469999, and
common difference d = 1.  We need n, the number of terms.
The formula for the nth term of an arithmetic series is

               an = a1 + (n - 1)d

          4469999 = 4440000 + (n - 1)(1)
 
Solve for n:

          4469999 = 4440000 + n - 1

          4469999 = 4439999 + n 

4469999 - 4439999 = n

            30000 = n

Edwin

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