How many 6-digit numbers can be formed from the digits 0, 1, 3, 5, 7 and 9 which are divisible by 10 and no digit is repeated?
120
Explanation
This question tests the application of permutations with constraints. The key insight is that a number divisible by 10 must end with 0. Since we need a 6-digit number using all 6 given digits without repetition, we must fix 0 at the unit's place and arrange the remaining 5 digits in the other positions. The multiplication principle from the context (as shown in Examples 3 and 10) applies here.
Solution Steps
Step 1: For a number to be divisible by 10, it must end with digit 0. Therefore, the unit's place must be 0.
Step 2: We have 6 digits: 0, 1, 3, 5, 7, 9. Since no digit is repeated and we need a 6-digit number, we must use all 6 digits exactly once.
Step 3: Fix 0 at the unit's place. Now we have 5 remaining digits (1, 3, 5, 7, 9) to fill the remaining 5 positions.
Step 4: The first position (leftmost) can be filled by any of the 5 remaining digits in 5 ways.
Step 5: The second position can be filled by any of the remaining 4 digits in 4 ways.
Step 6: Similarly, the third position can be filled in 3 ways, fourth position in 2 ways, and fifth position in 1 way.
Step 7: By multiplication principle, total number of 6-digit numbers = × × 1 = 120.