2024 AMC 10B Problem 20

Wherever an $L$ meets an $R$ in the side pattern, those two shoes must be mates. Thus an interior run cannot have length $1:$ its lone shoe would have to be the mate of both neighbors. With three $L$’s and three $R$’s, the only possible patterns are $LLLRRR,$ $RRRLLL,$ $LLRRRL,$ $LRRLLR,$ $LRRRLL,$ $RLLLRR,$ $RLLRRL,$ and $RRLLLR.$ For either one-switch pattern, choose the pair at the switch and order the remaining shoes, giving $3\cdot2\cdot2=12$ arrangements. Each of the other six patterns has $6$ assignments of the three pairs to its switches. Hence the total is $2\cdot12+6\cdot6=60.$ Therefore, the answer is A.