AMC8 2022

AMC8 2022 · Q14

AMC8 2022 · Q14. It mainly tests Permutations.

In how many ways can the letters in $\textbf{BEEKEEPER}$ be rearranged so that two or more $E$s do not appear together

字母 $\textbf{BEEKEEPER}$ 可以重新排列多少种方式，使得两个或更多 $E$s 不同时出现

(A) 1 1

(B) 4 4

(D) 24 24

(E) 120 120

Answer

Correct choice: (D)

正确答案：（D）

Solution

All valid arrangements of the letters must be of the form \[ \mathbf{E\underline{\quad}E\underline{\quad}E\underline{\quad}E\underline{\quad}E}. \] The problem is equivalent to counting the arrangements of $\mathbf{B},\ \mathbf{K},\ \mathbf{P}$, and $\mathbf{R}$ into the four blanks, in which there are $4! = \boxed{\textbf{(D)}\ 24}$ ways.

所有有效排列必须是这种形式： \[\mathbf{E\underline{\quad}E\underline{\quad}E\underline{\quad}E\underline{\quad}E}\] 问题等价于将$\mathbf{B},\ \mathbf{K},\ \mathbf{P},\ \mathbf{R}$填入四个空白，有$4! = \boxed{\textbf{(D)}\ 24}$种方式。

Topics

CP.002 Permutations