# Difference between revisions of "1983 AIME Problems/Problem 7"

## Problem

Twenty five of King Arthur's knights are seated at their customary round table. Three of them are chosen - all choices being equally likely - and are sent of to slay a troublesome dragon. Let $P$ be the probability that at least two of the three had been sitting next to each other. If $P$ is written as a fraction in lowest terms, what is the sum of the numerator and the denominator?

## Solution

### Solution 1

We can use Complementary counting by finding the probability that none are sitting next to each other and subtracting it from $1$.

Imagine the $22$ other (indistinguishable) people are already seated, and fixed into place.

We will place $A$, $B$, and $C$ with and without the restriction.

There are $22$ places to place $A$, followed by $21$ places to place $B$, and $20$ places to place $C$ after $A$ and $B$. Hence, there are $22\cdot21\cdot20$ ways to place $A, B, C$ in between these people with restrictions.

Without restrictions, there are $22$ places to place $A$, followed by $23$ places to place $B$, and $24$ places to place $C$ after $A$ and $B$. Hence, there are $22\cdot23\cdot24$ ways to place $A,B,C$ in between these people without restrictions.

Thus, the desired amount is $1-\frac{22\cdot21\cdot20}{22\cdot23\cdot24}=1-\frac{420}{552}=1-\frac{35}{46}=\frac{11}{46}$, and the answer is $11+46=\boxed{057}$.

### Solution 2

There are $(25-1)! = 24!$ configurations for the knights about the table.

There are ${3\choose 2} = 3$ ways to pick a pair of knights from the trio, and there are $2! = 2$ ways to determine which order they are seated. Since these two knights must be attached, we let them be a single entity, so there are $(24-1)! = 23!$ configurations for the entities.

However, this overcounts the instances in which the trio sits together; when all three knights sit together, then two of the pairs from the previous case are counted. However, we only want to count this as one case, so we need to subtract the number of instances in which the trio sits together (as a single entity). There are $3! = 6$ ways to determine their order, and there are $(23-1)! = 22!$ configurations.

Thus, the answer is $\frac{2 \times 3 \times 23! - 6 \times 22!}{24!} = \frac{11}{46}$, and the answer is $\boxed{057}$.

### Solution 3

Number the knights around the table 1-25. There are two possibilities: All three sit next to each other, or two sit next to each other and one is not sitting next to the other two.

Case 1: All three sit next to each other. In this case, you are picking $(1,2,3)$, $(2,3,4)$, $(4,5,6)$... $(25,1,2)$. This makes $25$ combinations.

Case 2: Like above, there are $25$ ways to pick the pair of knights sitting next to each other. Once a pair is picked, you cannot pick either of the two adjacent knights. (i.e. if you pick $(5,6)$, you may not pick 4 or 7). Thus, there are $25-4=21$ ways to pick the third knight, for a total of $25\cdot21$ combinations.

Thus, you have a total of $25 + (25\cdot21) = 25\cdot22$ allowable ways to pick the knights. The total number of ways to pick the knights is ${25\choose 3} = \frac{25\cdot24\cdot23}{3\cdot2\cdot1} = 25\cdot23\cdot4$.

The probability is $\frac{25\cdot22}{25\cdot23\cdot4} = \frac{11}{46}$, and the answer is $\boxed{057}$.

## See also

 1983 AIME (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
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