How many $7$-digit numbers are there with non-repeating digits?

Question

I have to find the amount of all $7$-digit numbers with non-repeating digits (every digit should be in the number only once). And also I have to find such a number with digits whose sum gives minimal possible number. I do not know how to approach that :/

Answer 1

Using the multiplication rule we get total number of 7 digit numbers as $9*9*8*7*6*5*4=544320$

The reasoning for this can be given as : For the first digit of the number we can select any of the nine of the nine digits from 1 to 9, for the second digit we can select any digit from 0 to 9 excluding the one which we have already placed in the first position . For the third place we have 8 options to choose from 0 to 9 excluding the digits we already placed on first and second positions. Hence the answer appears.

Also there will be $6*6*5*4*3*2*1=4320$ seven digit numbers having the digits $0,1,2,3,4,5,6$ giving minimum sum as 21. The reasoning is same as above.

Answer 2

Let us compute the number $a_n$ of $n$-different digits numbers which do not start with $0$. Clearly $a_n=0$ whenever $n>10$. Moreover $$ a_n=9 \cdot (9 \cdot 8 \cdots (11-n)) $$ if $n=1,\ldots,10$ (the second product is $1$ if $n=1$). The reason is that you have $9$ possibilities for the first digit on the left: all except $0$. Then, you will have $9$ possibilities for the second digit (including $0$), $8$ possibilities for the third one, etc.

Hence your answer is $$ a_1+a_2+\cdots +a_7. $$