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Find the total number of $9$ digit numbers which have all different digits.

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Hint: This is solved by using a number of permutations of $n$ distinct objects formula.

We have to make a $9$ digit number which have all different digits,
Now total number of digits$ = 10$
$i.e.{\text{ }}0,1,2,3,4,5,6,7,8,9$
And we require $9$ different numbers, thus $0$ can’t be placed first
$ \Rightarrow $ first place can be filled in $9$ ways
Now the ${2^{nd}}$ digit can be anything in the range $\left[ {0,9} \right]$ except for the one used for ${1^{st}}$ digit since we
do not want repetition of the digits.
$\therefore $ There are $9$ ways of filling the ${2^{nd}}$ digit of the number.
Similarly, ${3^{rd}}$ digit can be filled in 8 ways and so on….
Now total number of 9 digit numbers which have all different digits
$ = 9 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2$
$ = 9 \times 9!$

Note: In this question we simply use the number of permutations of $n$ different objects formula and also keeping the thing in mind that $0$ can’t be placed first, then we solve this question and we get our answer.
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