Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XI  >>  Math  >>  Permutations and Combinations
0 votes

The total number of 9 digit numbers which have all different digits is

$\begin{array}{1 1}(A)\;10!\\(B)\;9!\\(C)\;9.9!\\(D)\;10.10!\end{array} $

Can you answer this question?

1 Answer

0 votes
  • $n!=n(n-1)(n-2)(n-3)......(3)(2)(1)$
No of ways of filling first place =9
No of ways of filling remaining 8 place =$9\times 8\times 7\times 6\times 5\times 4\times 3\times 2$
$\Rightarrow 9!$
Total no of numbers =$9\times 9!$
Hence (C) is the correct answer.
answered May 19, 2014 by sreemathi.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App