Question

A digital lock is to be opened using all the digits (0 - 9). The code is a four digit number. If in unit and hundreds place only even digits are to be placed and in remaining digits only odd numbers are placed, find the number of wrong attempts that can be made.
1. 625
2. 624
3. 10 × 9 × 8 × 7) - 1
4. (10 × 9 × 8 × 7)
5. None of these

Answer 1

Correct Answer - Option 2 : 624

Concept:-

Basic concept of permutation.

Calculation:-

Unit and hundred place can be fixed with (2, 4, 6, 8, 0) and since nothing is mentioned about repetition. Number of ways of fixing the number = 5 × 5

Other digits have to be fixed with remaining odd digits, Number of ways = 5 × 5

Total number of combinations possible = 5 × 5 × 5 × 5 = 625

Since one of the combination is correct, number of wrong combinations = 625 - 1 = 624