Question

The access code for a car's security system consists of six digits. The first digit cannot be 5 and the last digit must be even. How many different codes are available? (Note that 0 is considered an even number).

Answer 1

We will use Fundamental Principal of Counting which states that : If there are n₁ ways to do first action , n₂ ways to do second action , n₃ ways to do third action and so on , then total no. of ways to do all 'm' actions together

= n₁*n₂*n₃* -------------*n_m  

So no. of different codes that are available

= (no. of ways we can fill in first place) * (no. of ways we can fill in second place) * (no. of ways to fill third place) * ----------------* (no. of ways to fill sixth place)

No. of ways to fill first place = we can use any digit from 0 to 9 except for 5 , so no. of options = 9

No. of ways to fill 2nd to 5 th place = we can use any digit from 0 to 9 , so no. of options = 10

No. of ways to fill 6 th place = we can use any even digit , so no. of options = 5

So, no. of different codes available = 9*10*10*10*10*5 = 450000