Find the total number of 9 digit number which have all different digits:

Concept:

PERMUTATION: Permutation:

Number of Permutations of ‘n’ things taken ‘r’ at a time:

\({\rm{p}}\left( {{\rm{n}},{\rm{\;r}}} \right) = \frac{{{\rm{n}}!}}{{\left( {{\rm{n}} - {\rm{r}}} \right)!}}\)

Number of Permutations of ‘n’ objects where there are n1 repeated items, n2 repeated items, ….. nk repeated items taken ‘r’ at a time:

\({\rm{p}}\left( {{\rm{n}},{\rm{\;r}}} \right) = \frac{{{\rm{n}}!}}{{{{\rm{n}}_1}!{{\rm{n}}_2}!{{\rm{n}}_3}! \ldots .{{\rm{n}}_{\rm{k}}}!}}\)

\(^np_0\ =\ 1\ and\ \ ^np_n\ =\ n!\)

Calculation:

A total number of 9 digit numbers having different digit is given by :

\({^{10}P_9} - {^9P_8} = \frac{10!}{1!}- \frac{9!}{1!}= 10! - 9!\)

= 10 × 9! - 9!

= 9! (10 - 1)

= 9! × 9

Hence, option 2 is correct.

Alternate MethodWe know that the total number of digits = 10 i.e.

0, 1, 2, 3, 4, 5, 6, 7, 8, 9

Out of which 0 can't be placed in the first place. 

Therefore, first place can be filled in 9 ways.

The rest 9 blanks with 9 digits in 9! ways 

Hence, a total number of ways = 9 × 9!