1. The sequence $\left\{a_{n}\right\}$ has 11 terms, satisfying
$$
a_{1}=0, a_{11}=4,
$$

and $\left|a_{k+1}-a_{k}\right|=1(k=1,2, \cdots, 10)$. The number of different sequences is ( ).
(A) 100
(B) 120
(C) 140
(D) 160
According to the problem, $a_{k+1}-a_{k}=1$ or -1.
In $a_{2}-a_{1}, a_{3}-a_{2}, \cdots, a_{11}-a_{10}$, suppose there are $x$ ones, then there are $10-x$ negative ones.
According to the problem, $4=x-(10-x) \Rightarrow x=7$.
Therefore, the number of sequences that meet the condition is $\mathrm{C}_{10}^{7}=120$. Hence, the answer is (B).