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 off 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 denominator?

Solution

Solution 1

We can use complementary counting, by finding the probability that none of the three knights are sitting next to each other and subtracting it from $1$ .

Imagine that 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 put $A$ , followed by $21$ places to put $B$ , and $20$ places to put $C$ 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 put $A$ , followed by $23$ places to put $B$ , and $24$ places to put $C$ 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 probability 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 (since configurations that are derived from each other simply by a rotation are really the same, and should not be counted multiple times).

There are ${3\choose 2} = 3$ ways to pick a pair of knights from the trio, and there are $2! = 2$ ways to determine in which order they are seated. Since these two knights must be together, 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 required probability 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 from $1$ to $25$ . The total number of ways to pick the knights is ${25\choose 3} = \frac{25\cdot24\cdot23}{3\cdot2\cdot1} = 25\cdot23\cdot4$ There are two possibilities: either 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 $(1,2,3)$ , $(2,3,4)$ , $(3,4,5)$ , $(4,5,6)$ ,..., $(23,24,25)$ , $(24,25,1)$ , $(25,1,2)$ . 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. (For example, 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 probability is $\frac{25\cdot22}{25\cdot23\cdot4} = \frac{11}{46}$ , and the answer is $\boxed{057}$ .

Solution 4

Pick an arbitrary spot for the first knight. Then pick spots for the next two knights in order.

Case 1: The second knight sits next to the first knight. There are $2$ possible places for this out of $24$ , so the probability of this is $\frac{1}{12}$ . We do not need to consider the third knight.

Case 2: The second knight sits two spaces apart from the first knight. There are $2$ possible places for this out of $24$ , so the probability is $\frac{1}{12}$ . Then there are $3$ places out of a remaining $23$ for the third knight to sit, so the total probability for this case is $\frac{1}{12} \times \frac{3}{23}$ .

Case 3: The second knight sits three or more spaces apart from the first knight. There are $20$ possible places for this out of $24$ , so the probability is $\frac{5}{6}$ . Then there are $4$ places to put the last knight out of $23$ , so the total probability for this case is $\frac{5}{6}\times\frac{4}{23}$ .

Now we add the probabilities to get the total:

$\frac{1}{12}+\frac{1}{12}\times\frac{3}{23}+\frac{5}{6}\times\frac{4}{23}=\frac{1}{12}\times\frac{1}{23}\left(23+3+40\right)=\frac{66}{12\times 23}$ $=\frac{6\times 11}{6\times 2 \times 23}=\frac{11}{46}$ so the answer is $\boxed{057}$ .

Solution 5

We simplify this problem by using complementary counting and fixing one knight in place. Then, either a knight can sit two spaces apart from the fixed knight, or a knight can sit more than two spaces apart from the fixed knight. The probability is then $\frac{24\left(23\right)-\left[2\left(20\right)+20\left(19\right)\right]}{24\left(23\right)}=\frac{11}{46}$ , so the answer is $11+46=\boxed{057}$ .

1983 AIME (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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