Question

Question 4

Part 1
In a different plan for area​ codes, the first digit could be any number from 1 through​ 6, the second digit was either 6 ,7 or 8​, and the third digit could be any number except 1 or 2. With this​ plan, how many different area codes are​ possible?

There are _____possible area codes.

Answer 1

144

Explanation:

The code includes 3 digits in total:

First digit: 6 choices (1; 2; 3; 4; 5; 6)

Second digit: 3 choices (6; 7; 8)

Third digit: 8 choices (0; 3; 4; 5; 6; 7; 8; 9)

Use the combinatorial multiplication rule to find the amount of possible area codes (it hasn't been said that the all digits must be different):

6 × 3 × 8 = 144

We do not divide by the factorial because the order of compiling the codes is important

I don't know if I got it right, though...