AMC10 2024 B
AMC10 2024 B · Q20
AMC10 2024 B · Q20. It mainly tests Basic counting (rules of product/sum), Casework.
Three different pairs of shoes are placed in a row so that no left shoe is next to a
right shoe from a different pair. In how many ways can these six shoes be lined up?
三双不同的鞋排成一排，使得没有左鞋与不同双的右鞋相邻。这六只鞋有多少种排列方式？
(A)
60
60
(B)
72
72
(C)
90
90
(D)
108
108
(E)
120
120
Answer
Correct choice: (A)
正确答案：（A）
Solution
Let $A_R, A_L, B_R, B_L, C_R, C_L$ denote the shoes.
There are $6$ ways to choose the first shoe. WLOG, assume it is $A_R$. We have $A_R,$ __, __, __, __, __.
$~~~~~$ Case $1$: The next shoe in line is $A_L$. We have $A_R, A_L,$ __, __, __, __. Now, the next shoe in line must be either $B_L$ or $C_L$. There are $2$ ways to choose which one, but assume WLOG that it is $B_L$. We have $A_R, A_L, B_L,$ __, __, __.
$~~~~~ ~~~~~$ Subcase $1$: The next shoe in line is $B_R$. We have $A_R, A_L, B_L, B_R,$ __, __. The only way to finish is $A_R, A_L, B_L, B_R, C_R, C_L$.
$~~~~~ ~~~~~$ Subcase $2$: The next shoe in line is $C_L$. We have $A_R, A_L, B_L, C_L,$ __, __. The only way to finish is $A_R, A_L, B_L, C_L, C_R, B_R$.
$~~~~~$ In total, this case has $(6)(2)(1 + 1) = 24$ orderings.
$~~~~~$ Case $2$: The next shoe in line is either $B_R$ or $C_R$. There are $2$ ways to choose which one, but assume WLOG that it is $B_R$. We have $A_R, B_R,$ __, __, __, __.
$~~~~~ ~~~~~$ Subcase $1$: The next shoe is $B_L$. We have $A_R, B_R, B_L,$ __, __, __.
$~~~~~ ~~~~~ ~~~~~$ Sub-subcase $1$: The next shoe in line is $A_L$. We have $A_R, B_R, B_L, A_L,$ __, __. The only way to finish is $A_R, B_R, B_L, A_L, C_L, C_R$.
$~~~~~ ~~~~~ ~~~~~$ Sub-subcase $2$: The next shoe in line is $C_L$. We have $A_R, B_R, B_L, C_L,$ __, __. The remaining shoes are $C_R$ and $A_L$, but these shoes cannot be next to each other, so this sub-subcase is impossible.
$~~~~~ ~~~~~$ Subcase $2$: The next shoe is $C_R$. We have $A_R, B_R, C_R,$ __, __, __. The next shoe in line must be $C_L$, so we have $A_R, B_R, C_R, C_L,$ __, __. There are $2$ ways to finish, which are $A_R, B_R, C_R, C_L, A_L, B_L$ and $A_R, B_R, C_R, C_L, B_L, A_L$.
$~~~~~$ In total, this case has $(6)(2)(1 + 2) = 36$ orderings.
Our final answer is $24 + 36 = \boxed{\textbf{(A) } 60}$
设$A_R, A_L, B_R, B_L, C_R, C_L$表示鞋子。
第一只鞋有$6$种选择。假设是$A_R$。$A_R,$ __, __, __, __, __。
~~情况$1$：第二只是$A_L$。$A_R, A_L,$ __, __, __, __。第三必须是$B_L$或$C_L$，$2$种选择，假设是$B_L$。$A_R, A_L, B_L,$ __, __, __。
~~~~子情况$1$：第四是$B_R$。$A_R, A_L, B_L, B_R,$ __, __。只剩$A_R, A_L, B_L, B_R, C_R, C_L$一种。
~~~~子情况$2$：第四是$C_L$。$A_R, A_L, B_L, C_L,$ __, __。只剩$A_R, A_L, B_L, C_L, C_R, B_R$一种。
~~~~本情况共$(6)(2)(1 + 1) = 24$种。
~~情况$2$：第二是$B_R$或$C_R$，$2$种，假设$B_R$。$A_R, B_R,$ __, __, __, __。
~~~~子情况$1$：第三是$B_L$。$A_R, B_R, B_L,$ __, __, __。
~~~~~子子情况$1$：第四是$A_L$。$A_R, B_R, B_L, A_L,$ __, __。只剩$A_R, B_R, B_L, A_L, C_L, C_R$一种。
~~~~~子子情况$2$：第四是$C_L$。剩余$C_R,A_L$不能相邻，不可能。
~~~~子情况$2$：第三是$C_R$。$A_R, B_R, C_R,$ __, __, __。第四必须$C_L$，$A_R, B_R, C_R, C_L,$ __, __。后两位置$2$种：$A_R, B_R, C_R, C_L, A_L, B_L$和$A_R, B_R, C_R, C_L, B_L, A_L$。
~~~~本情况共$(6)(2)(1 + 2) = 36$种。
总计$24 + 36 = \boxed{\textbf{(A) } 60}$
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