34. Using the digits $1,2,3,4,5,6,7,8$, we can form 8 ! $(=40320) 8$-digit numbers in which the eight digits are all distinct. For $1 \leq k \leq 40320$, let $a_{k}$ denote the $k$ th number if these numbers are arranged in increasing order:
$$
12345678,12345687,12345768, \ldots, 87654321 ;
$$
that is, $a_{1}=12345678, a_{2}=12345687, \ldots, a_{40320}=87654321$. Find $a_{2099}-a_{2001}$.
34. Answer: 441
First we determine $a_{2005}$ and $a_{2009}$. Suppose that $a_{2005}=\overline{x_{1} x_{2} x_{3} x_{4} x_{5} x_{6} x_{7} x_{4}}$, where the $x_{i}$ 's are distinct digits in $\{1,2,3,4,5,6,7,8\}$.
Let $A=\left\{a_{k}: k=1,2, \ldots, 40320\right\}$.
Since $7!=5040>2008$, we deduce that $x_{1}=1$, as there are more than 2008 numbers in $A$ such that the first digit is 1 .
As $2 \times 6!<2008<3 \times 6$ !, we have $x_{2}=4$, as there are less than 2008 numbers in A such that the first digit is 1 and the second digit is 2 or 3 , but there are more than 2008 numbers in $A$ such that the first digit is 1 and the second digit is 2,3 or 4. Similarly, since $2 \times 6!+4 \times 5!<2008<2 \times 6!+5 \times 5$ !, we see that the third digit $x_{3}$ is 7 . By repeating the argument and using the inequalities
$$
\begin{array}{l}
2 \times 6!+4 \times 5!+3 \times 4!<2008<2 \times 6!+4 \times 5!+4 \times 4!\text { and } \\
2004=2 \times 6!+4 \times 5!+3 \times 4!+2 \times 3!<2008<2 \times 6!+4 \times 5!+3 \times 4!+3 \times 3!\text {, }
\end{array}
$$
we obtain $x_{4}=6, x_{5}=5$. Note also that among the numbers in $A$ of the form $1476^{* * * *}$, the digit 5 first appears as the fifth digit in $a_{\text {20os }}$ if the numbers are arranged in increasing order. Consequently, as the last three digits are 2,3 and 8 , we must have $a_{2 \cos }=14765238$. It follows that $a_{2000}=14765283$,
$$
\begin{array}{l}
a_{2007}=14765328, a_{2008}=14765382, \text { and } a_{2009}=14765823 \text {. Hence } \\
a_{2009}-a_{2008}=14765823-14765382=441 .
\end{array}
$$