Permutations Practice Test 2

Since 5 does not occur in 1000, we have to count the number of times 5 occurs when we list the integers from 1 to 999. Any number between 1 and 999 is of the form xyz, 0 ≤ x, y, z ≤ 9.
The number in which 5 occurs exactly once $={(}^3{\mathit{C}}_1)\times 9\times 9=243$.
The numbers in which 5 occurs exactly twice $={(}^3{\mathit{C}}_2\times 9)=27$.
The numbers in which 5 occurs in all three digits = 1.
Hence, the number of times 5 occurs is 1 × 243 + 2 × 27 + 3 × 1 = 300