smc

We say that a game state is an N-position if it is winning for the next player (the player to move), and a P-position if it is winning for the other player. We are trying to find which of the given states is a P-position. First we note that symmetrical positions are P-positions, as the second player can win by mirroring the first player's moves. It follows that $(6,1,1)$ is an N-position, since we can win by moving to $(2,2,1,1)$; this rules out $\text{(A)}$. We next look at $(6,2,1)$. The possible next states are $$(6,2),(6,1,1),(6,1),(5,2,1),(4,2,1,1),(4,2,1),(3,2,2,1),(3,2,1,1),(2,2,2,1).$$ None of these are symmetrical, so we might reasonably suspect that they are all N-positions. Indeed, it just so happens that for all of these states except $(6,2)$ and $(6,1)$, we can win by moving to $(2,2,1,1)$; it remains to check that $(6,2)$ and $(6,1)$ are N-positions. To save ourselves work, it would be nice if we could find a single P-position directly reachable from both $(6,2)$ and $(6,1)$. We notice that $(3,2,1)$ is directly reachable from both states, so it would suffice to show that $(3,2,1)$ is a P-position. Indeed, the possible next states are $$(3,2),(3,1,1),(3,1),(2,2,1),(2,1,1,1),(2,1,1),$$ which allow for the following refutations: $$\begin{array}{rlrlrlrlrl}& (3,2)\to (2,2),& & & & (3,1,1)\to (1,1,1,1),& & & & (3,1)\to (1,1),\\ & (2,2,1)\to (2,2),& & & & (2,1,1,1)\to (1,1,1,1),& & & & (2,1,1)\to (1,1).\end{array}$$ Hence, $(3,2,1)$ is a P-position, so $(6,2)$ and $(6,1)$ are both N-positions, along with all other possible next states from $(6,2,1)$ as noted before. Thus, $(6,2,1)$ is a P-position, so our answer is $\boxed{(6,2,1)}$. (For completeness, we could also rule out $\text{(C)}$, $\text{(D)}$ and $\text{(E)}$ as in Solution 2.)