MCQ
The number of nine digit numbers that can be formed with different digits is:
- A9.8!
- B8.9!
- C9.9!
- D10!
Solution:
The first digit is one of (1, 2, 3, 4, 5, 6, 7, 8, 9) (i.e., all expect 0).
So, the no of ways $=9$
The remaining 8 digits are to be taken from the remaining 9 numbers.
So, the no of ways $= \ ^9\text{P}_8$
So, total no. of ways $=9\times\ ^9\text{P}_8$
$=9\times9!$
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