A) \[9\times 9\ !\]
B) \[9\ !\]
C) 10!
D) None of these
Correct Answer: A
Solution :
There are 10 digits in all viz. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The required 9 digit numbers = (Total number of 9 digit numbers including those numbers which have 0 at the first place) - (Total number of those 9 digit numbers which have 0 at the first place) \[={{\,}^{10}}{{P}_{9}}-{{\,}^{9}}{{P}_{8}}=\frac{10\,!}{1\,!}-\frac{9\,!}{1\,!}=10\,!\,\,-9\,!\] = \[(10-1)\,9\,!=9.9\,!\].
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