AIME 2014 II · 第 9 题

Solution 1 (Casework)

We know that a subset with less than $3$ chairs cannot contain $3$ adjacent chairs. There are only $10$ sets of $3$ chairs so that they are all $3$ adjacent. There are $10$ subsets of $4$ chairs where all $4$ are adjacent, and $10 \cdot 5$ or $50$ where there are only $3.$ If there are $5$ chairs, $10$ have all $5$ adjacent, $10 \cdot 4$ or $40$ have $4$ adjacent, and $10 \cdot {5\choose 2}$ or $100$ have $3$ adjacent. With $6$ chairs in the subset, $10$ have all $6$ adjacent, $10(3)$ or $30$ have $5$ adjacent, $10 \cdot {4\choose2}$ or $60$ have $4$ adjacent, $\frac{10 \cdot 3}{2}$ or $15$ have $2$ groups of $3$ adjacent chairs, and $10 \cdot \left({5\choose2} - 3\right)$ or $70$ have $1$ group of $3$ adjacent chairs. All possible subsets with more than $6$ chairs have at least $1$ group of $3$ adjacent chairs, so we add ${10\choose7}$ or $120$, ${10\choose8}$ or $45$, ${10\choose9}$ or $10$, and ${10\choose10}$ or $1.$ Adding, we get $10 + 10 + 50 + 10 + 40 + 100 + 10 + 30 + 60 + 15 + 70 + 120 + 45 + 10 + 1 = \boxed{581}.$

Solution 2 (PIE)

Starting with small cases, we see that four chairs give $4 + 1 = 5$, five chairs give $5 + 5 + 1 = 11$, and six chairs give $6 + 6 + 6 + 6 + 1 = 25.$ Thus, n chairs should give $n 2^{n-4} + 1$, as confirmed above. This claim can be verified by the principle of inclusion-exclusion: there are $n 2^{n-3}$ ways to arrange $3$ adjacent chairs, but then we subtract $n 2^{n-4}$ ways to arrange $4.$ Finally, we add $1$ to account for the full subset of chairs. Thus, for $n = 10$ we get a first count of $641.$

However, we overcount cases in which there are two distinct groups of three or more chairs. We have $5$ cases for two groups of $3$ directly opposite each other, $5$ for two groups of four, $20$ for two groups of $3$ not symmetrically opposite, $20$ for a group of $3$ and a group of $4$, and $10$ for a group of $3$ and a group of $5.$ Thus, we have $641 - 60 = \boxed{581}$.

Solution 3 (Complementary Counting and Recursion)

It is possible to use recursion to count the complement. Number the chairs $1, 2, 3, ..., 10.$ If chair $1$ is not occupied, then we have a line of $9$ chairs such that there is no consecutive group of three. If chair $1$ is occupied, then we split into more cases. If chairs $2$ and $10$ are empty, then we have a line of $7.$ If chair $2$ is empty but chair $10$ is occupied, then we have a line of $6$ chairs (because chair $9$ cannot be occupied); this is similar to when chair $2$ is occupied and chair $10$ is empty. Finally, chairs $2$ and $10$ cannot be simultaneously occupied. Thus, we have reduced the problem down to computing $T_9 + T_7 + 2T_6$, where $T_n$ counts the ways to select a subset of chairs $\textit{in a line}$ from a group of n chairs such that there is no group of $3$ chairs in a row.

Now, we notice that $T_n = T_{n-1} + T_{n-2} + T_{n-3}$ (representing the cases when the first, second, and/or third chair is unoccupied). Also, $T_0 = 1, T_1 = 2, T_2 = 4, T_3 = 7$, and hence $T_4 = 13, T_5 = 24, T_6 = 44, T_7 = 81, T_8 = 149, T_9 = 274$. Now we know the complement is $274 + 81 + 88 = 443$, and subtracting from $2^{10} = 1024$ gives $1024 - 443 = \boxed{581}$.

Solution 4

Let's calculate the complement. As mentioned in solution $3$, the number of ways to have a subset $n$ chairs in a line with no 3 consectuive chairs satisfies $T_n = T_{n-1} + T_{n-2} + T_{n-3}$. Setting $T_{1} = 2, T_{2} = 4$, and $T_{3} = 7$, we get that $T_{10} = 504$. Since this is in a line and not a circle, we must subtract the cases that would include 3 consecutive chairs if the endpoints of the line were put together. If chairs $1$, $2$ and $10$ are in the subset, that would not work. The same goes for if chairs $1$, $9$ and $10$ were in the subset. If chairs $1$, $2$ and $10$ are in the set then chair $3$ must not be in the set. However, chair $9$ could be in or not in the set because we only want to count what cases where no $3$ chairs are consecutive in the line but there would be consecutive chairs in a circle. If chair $9$ is not included, there are $T_{5} = 24$ ways. If chair $9$ is in the set then there are $T_{4} = 13$ ways. So we must subtract $2 \cdot (24 + 13) = 74$ However we are counting the case where chairs $1$, $2$, $9$ and $10$ are included twice. So we only have to subtract $74 - 13 = 61$. $504 - 61 = 443$. This entire time we were calculating the complement so $2^{10} - 443 = \fbox{581}$

~sdfgfjh