Question 1195492
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How many 5-digit {{{highlight(positive_integer)}}} numbers can be formed using {{{highlight(cross(numbers))}}} <U>digits</U> 0, 1, 2, …9, 
such that the first digit must not be 9 and repetition is not allowed
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<pre>
Any of 8 digits  {1,2,3, . . . ,8}  can be in the first, most-left position 
    (everything except of 0 or 9),                                giving 8 possible options.


Any of remaining  10-1 = 9  digits  can be in the next position,  giving 9 possible options.

Any of remaining   9-1 = 8  digits  can be in the next position,  giving 8 possible options.

Any of remaining   8-1 = 7  digits  can be in the next position,  giving 7 possible options.

Any of remaining   7-1 = 6  digits  can be in the next (= the last) position,  giving 6 possible options.


The total number of 5-digit numbers satisfying the imposed conditions is  8*9*8*7*6 = 24192.    <U>ANSWER</U>
</pre>

Solved.



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To Alan:


<pre>
    Alan, in such problems, the leading digit 0 is PROHIBITED by default.
</pre>