The total number of 9-digit numbers of different digits is :a)10(9!)b)...
The first place from the left can' be filled in 9 ways (any one except 0).
The other eight places can be filled by the remaining 9 digits in 9P8 ways.
∴ the number of 9 -digit numbers = 9 x 9P8.
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The total number of 9-digit numbers of different digits is :a)10(9!)b)...
The first place from the left can' be filled in 9 ways (any one except 0).
The other eight places can be filled by the remaining 9 digits in 9P8 ways.
∴ the number of 9 -digit numbers = 9 x 9P8.
The total number of 9-digit numbers of different digits is :a)10(9!)b)...
- Explanation:
The total number of 9-digit numbers of different digits can be calculated by considering each digit's possibilities in each position of the number.
- Calculating the possibilities:
- For the first digit of the number, we have 9 choices (1-9).
- For the second digit, we have 9 choices (we can use any digit except for the one used in the first position).
- Similarly, for the third digit, we have 8 choices, for the fourth digit, we have 7 choices, and so on until the ninth digit where we have 1 choice left.
- Using the Fundamental Counting Principle:
- By using the Fundamental Counting Principle, we multiply the number of choices for each position to find the total number of 9-digit numbers of different digits:
- Total = 9 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 = 9!
- Conclusion:
- Therefore, the total number of 9-digit numbers of different digits is 9 times the factorial of 9, which is represented as 9(9!). Hence, the correct answer is option 'C'.