Question

In a different plan for area​ codes, the first digit could be any number from 0 through 6 , the second digit was either 7 or 8​, and the third digit could be any number except 0 or 6. or 6. With this​ plan, how many different area codes are​ possible?

Answer 1

112 different area codes can be made.

Solution:

Given, In a different plan for area codes, the first digit could be any number from 0 through 6 ,  

So, total 7 digits are possible for 1st place (i.e. 0 1 2 3 4 5 6)

The second digit was either 7 or 8,  

So, total 2 digits are possible for 2nd place (i.e. 7 8)

And the third digit could be any number except 0 or 6.  

So, total possible digits are 8 for 3rd place (i.e. 1 2 3 4 5 7 8 9)

With this plan, we have to find how many different area codes are possible?

$\text { total possible combinations }=7 \text { possibilities for } 1 \text { st place } \times 2 \text { for } 2 \text { nd place } \times 8 \text { for } 3 \text { rd place }$

$\text { Total possible area codes }=7 \times 2 \times 8=14 \times 8=112$

Hence, 112 different area codes can be made.