<p>Find the total number of 9-digit numbers which have all different digits.</p>
Step-by-Step Solution
Key Concept: Count all permutations of 9 digits from 10, then subtract cases where 0 is in the first position (invalid).
<p><strong>Solution:</strong></p><p>The digits available are 10: (0, 1, 2, 3, 4, 5, 6, 7, 8, 9)</p><p>Total number of 9-digit arrangements from 10 digits = <span style="math">$^{10}P_9$</span></p><p>However, numbers having 0 at the first place are invalid = <span style="math">$^9P_8$</span></p><p>Required number of valid 9-digit numbers = <span style="math">$^{10}P_9 - \,^9P_8$</span></p><p><span style="math">$= 10 \times \,^9P_8 - \,^9P_8$</span></p><p><span style="math">$= 9 \times \,^9P_8 = 9 \times 9!$</span></p>
Correct Answer: 9 × 9!