MCQ
Choose the correct answer. The total number of $9$ digit numbers which have all different digits is.
  • A
    $10!$
  • B
    $9 !$
  • $9 \times 9!$
  • D
    $10\times 10!$

Answer

Correct option: C.
$9 \times 9!$
We have to form $9-$digit number which has all different digit.
First digit from the left can be filled in $9$ ways $($excluding $'0\ ').$
Now nine digits are left including $'O\ '.$
So remaining eight places can be filled with these nine digits in ${ }^9 P_S$ ways.
So, total number of numbers $=9 \times{ }^9 \mathrm{P}_8=9 \times 9$ !

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free