10. Given a sequence $\left\{a_{n}\right\}$ with 100 terms, satisfying $a_{1}=0, a_{100}=475$,
and $\left|a_{k+1}-a_{k}\right|=5(k=1,2, \cdots, 99)$.
Then the number of different sequences that meet the conditions is $\qquad$.
10.4 851 .

Notice that, $a_{k+1}-a_{k}=5$ or -5 ,
$$
\begin{array}{l}
a_{10}=\left(a_{100}-a_{99}\right)+\left(a_{99}-a_{98}\right)+\cdots+\left(a_{2}-a_{1}\right) \\
=475 .
\end{array}
$$

Let there be $x$ instances of 5 and $99-x$ instances of -5 among the 99 differences.
Thus, $5 x+(-5)(99-x)=475 \Rightarrow x=97$.
Therefore, in the sequence of 99 differences $a_{k+1}-a_{k}$ $(k=1,2, \cdots, 99)$, there are 97 instances of 5 and 2 instances of -5.

Since each permutation of these 97 instances of 5 and 2 instances of -5 uniquely corresponds to a sequence that meets the conditions, the number of such sequences is $\mathrm{C}_{99}^{2}=4851$.