Question

由 $0$ 和 $1$ 组成的长度为 $19$ 的序列中，有多少个以 $0$ 开头、以 $0$ 结尾，且既不含两个连续的 $0$ ，也不含三个连续的 $1$ ？

How many sequences of $0$ s and $1$ s of length $19$ are there that begin with a $0,$ end with a $0,$ contain no two consecutive $0$ s, and contain no three consecutive $1$ s?

$55$

$60$

$65$

$70$

$75$

答案：C

知识点：有限制的排列分拆与有序分拆组合

难度评级：1770

解答：

序列先以一个 $0$ 开头，之后由若干个 $110$ 块和 $10$ 块依次组成，每个块都接在一个 $0$ 后面。

设 $110$ 块有 $x$ 个， $10$ 块有 $y$ 个，则序列的总长度为所以可能的有序对只有对每一组解，排列这 $x+y$ 个块时，需要选出 $x$ 个位置放置 $110$ 块，因此从 $x+y$ 个位置中选择的方法数为 $3x+2y+1=19,$ $3x+2y=18.$ $\begin{aligned} (x,y)&=(6,0),(4,3),\\ &\quad(2,6),(0,9). \end{aligned}$ $\binom{x+y}x,$

因此总数为 $\binom 66 + \binom 74 + \binom 82 + \binom 90$ $=1+35+28+1$ $=65.$

所以正确答案是 C。

Our sequence starts with a $0$ then has sequences of $110$ and $10$ in some order, where they each come after a $0.$

Let the number of $110$ be $x$ and let the number of $10$ be $y.$ Then the number of terms in the sequence is $3x+2y+1=19,$ making $3x+2y=18.$ The possible ordered pairs are $\begin{aligned} (x,y)&=(6,0),(4,3),\\ &\quad(2,6),(0,9). \end{aligned}$ Then, the number of ways to order the $x+y$ blocks is $\binom{x+y}x,$ since we choose which $x$ of the $x+y$ positions hold a $110$ block.

Therefore, the total number of ways is $\binom 66 + \binom 74 + \binom 82 + \binom 90$ $=1+35+28+1$ $=65.$

Thus, the answer is C .

2019 AMC 10B 第 25 题

答案：C

难度评级：1770

解答：

其他年份的第 25 题