Question

In a different plan for area codes, the first digit could be any number from 3 through 8, the second digit was either 2,3, or 4 and the third digit could be any number except 6,7, or 8. With this plan, how many different area codes are possible?

Answer 1

We count the total number of possibilities for the area codes in the next manner:

1. For the first digit of the codes, we have six options, these are; 3,4,5,6,7,8

2. For the second digit of the codes, three options\; 2,3,4

3. For the third digit of the codes with can put any digit number except 6,7 or 8, that restriction let us the following possibilities:0,1,2,3,4,5,9

We will use the multiplicative rule to count the total number of possible codes with this restrictions, that is we will apply a rule of the form:

`total\text{ N' of codes=possibilities for the first digit}*(possibilities\text{ for the second digid\rparen}*(possibilities\text{ for the third digid\rparen}`

In our the specific case

`Total\text{ N' od codes=6}*3*7=126`

Therefore, we conclude that with these restriction, the total number of codes is 126