Use the fundamental counting principle to solve the problem. How many different codes of 4 digits are possible if the first digit must be 3, 4, or 5 and if the code may not end in 0?

Answer and Explanation:

Given,

4 digits and 10 digits.

Firstly, we have to find that when it repeats the digit then

First position code would be filled in by Three methods that is 3, 4, or 5.Second position code would be filled in by Ten methods.Third position code would also be filled in by Ten methods.Fourth position code would be filled in by 9 way but 0 is rejected. So, the total numbers of methods in the following case is .

Then, we have to find that when it not repeats the digit then

First position code would be filled in by Three methods that is 3, 4, or 5.Second position code would be filled in by Nine methods in which one is reserved earlier.Third position code would be filled in by Eight methods in which two are reserved earlier.Fourth position code would be filled in by Six methods in which three are reserved earlier.So, the total numbers of methods in the following case is

Binary code is the right answer