Answer (A): There are $\binom{6}{3} = 20$ arrangements of the letters LLLRRR representing the positions of 3 left shoes and 3 right shoes in the row of 6 shoes. Of the 20, any sequence containing LRL or RLR will violate the condition given in the problem. There are 8 arrangements that avoid these two sequences. Call a pair of shoes matched if the 2 shoes in the pair are next to each other. There are three sets of possibilities.
* LLLRRR and RRRLLL: In these two cases, only one pair of shoes is matched. There are 3 choices for that pair, and there are $2 \cdot 2 = 4$ ways to place the other 4 shoes for a total of $6 \cdot 4 = 24$ arrangements for this case.
* LRRRLL, LLRRRL, RLLLR, and RRLLLR: In each of these four cases, there are $3! = 6$ ways to place the left shoes, but then there is a unique way to place the right shoes, for a total of $6 \cdot 4 = 24$ arrangements in this case.
* LRRLLR and RLLRRL: In these two cases, all three pairs of shoes are matched, so, in each case, there are $3! = 6$ ways to place the shoes. This gives a total of $6 \cdot 2 = 12$ arrangements.
Thus there are $24 + 24 + 12 = 60$ arrangements satisfying the conditions of the problem.

Label the shoes $L_i$ and $R_i$ for $i = 1, 2, 3$. By symmetry it suffices to count the number of arrangements with $L_1$ first in line and multiply by 6. There are 2 choices for a left shoe coming next, say $L_2$, after which the only continuations are $L_3 R_3 R_j R_{3-j}$ for $j = 1$ or $2$, or $R_2 R_1 R_3 L_3$; this gives $2 \cdot (2 + 1) = 6$ arrangements. Otherwise $R_1$ comes second, followed by either $R_k R_{5-k} L_{5-k} L_k$ or $R_k L_k L_{5-k} R_{5-k}$ for $k = 2$ or $3$, another $2 + 2 = 4$ arrangements. This gives a total of 10, so there are $6 \cdot 10 = 60$ ways to line up the six shoes.