2025 AMC 10A Problems/Problem 14: Difference between revisions

Revision as of 20:06, 6 November 2025

Six chairs are arranged around a round table. Two students and two teachers randomly select four of the chairs to sit in. What is the probability that the two students will sit in two adjacent chairs and the two teachers will also sit in two adjacent chairs?

$\textbf{(A) } \frac 16 \qquad \textbf{(B) } \frac 15 \qquad \textbf{(C) } \frac 29 \qquad \textbf{(D) } \frac 3{13} \qquad \textbf{(E) } \frac 14$

Solution 1

First, we count the number of desired outcomes ( $2$ students sit together, $2$ teachers sit together). We'll start by treating the $2$ students as a block and doing the same for the teachers. Hence, we are seating $2$ blocks in round table with $4$ seats. It doesn't matter where we sit the student block since we can just rotate the table so they're at the top. After the student block is seated, there are $3$ open seats for the teacher block. Since both students and both teachers can switch seats in their blocks, there are $3\cdot2^2=12$ desired outcomes.

For the total outcomes, we first place one of the students, say Jimmy. It doesn't matter where we place Jimmy, as we can always rotate the table so he's at the top. Then, there are $5\cdot4\cdot3$ ways to place the other student and the two teachers. There are hence $5\cdot4\cdot3=60$ total outcomes. The answer is $\boxed{\text{(B) }\dfrac{1}{5}}.$

~Tacos_are_yummy_1

Solution 2

There are $6$ ways to select the first pair of adjacent chairs and $3$ ways to select the next. There are $2!$ ways to permute the students and teachers amongst themselves and $\binom{6}{4} \cdot 4!$ total ways they can sit down, giving us $\frac{6 \cdot 3 \cdot (2!)^2}{\binom{6}{4} \cdot 4!} = \boxed{\text{(B) }\dfrac{1}{5}}.$

~grogg007

Solution 3

First, we count the total number of outcomes: there are $6$ choices for student $1$ , $5$ for student $2$ , $4$ for teacher $1$ , and $3$ for teacher $2$ to make a total of $6 * 5* 4 * 3 = 360$ total outcomes.

Next, we count the desired ones. There are $6$ choices for student $1$ , $2$ for student $2$ , and $3$ for the two teachers, however, since they are different and have order, there is $6 * 2 * 3 * 2! = 72$ desired outcomes.

Therefore, our answer is $\frac{72}{360} = \boxed{\frac{1}{5}}$

~iiiiiizh

Solution 4

There are six seats for a student to choose, Now, the next student can sit to the left or right to first, giving a $\frac{2}{5}$ probability for the two students to sit next to each other. As for the teachers, the first one has four seats to choose from, but two of them will be next to the students. As for this case you get $\frac{1}{2}$ , and then multiply by $\frac{1}{3}$ , which is the chance that the second teacher sits next to the first. Applying the same process for the other two seats gives ${\frac{1}{2}}\cdot{\frac{2}{3}}=\frac{1}{3}$ . Adding the two probabilities gives $\frac{1}{2}$ , and mulitplying by the $\frac{2}{5}$ we got earlier from the students we get a final answer of $\boxed{\frac{1}{5}}$ .

~Avy11

Video Solution (Fast and Easy)

https://youtu.be/sLhVhTM9HmA?si=3CeX_qoB7_GHdbHs ~ Pi Academy

Video Solution

https://youtu.be/gWSZeCKrOfU

~MK

@@ Line 32: / Line 32: @@
 There are six seats for a student to choose, Now, the next student can sit to the left or right to first, giving a <imath>\frac{2}{5}</imath> probability for the two students to sit next to each other. As for the teachers, the first one has four seats to choose from, but two of them will be next to the students. As for this case you get <imath>\frac{1}{2}</imath>, and then multiply by <imath>\frac{1}{3}</imath>, which is the chance that the second teacher sits next to the first. Applying the same process for the other two seats gives <imath>{\frac{1}{2}}\cdot{\frac{2}{3}}=\frac{1}{3}</imath>. Adding the two probabilities gives <imath>\frac{1}{2}</imath>, and mulitplying by the <imath>\frac{2}{5}</imath> we got earlier from the students we get a final answer of <imath>\boxed{\frac{1}{5}}</imath>.
+~Avy11
 == Video Solution (Fast and Easy) ==

2025 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
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