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

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==Solution 2==  
==Solution 2==  


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


~[[User:grogg007|grogg007]]
~[[User:grogg007|grogg007]]

Revision as of 12:32, 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 $\frac{6 \cdot 3 \cdot (2!)^2}{\binom{6}{4} \cdot 4!} = \boxed{\frac{1}{5}}.$

~grogg007

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