2025 AMC 12A Problems/Problem 6: Difference between revisions
Created page with "== Problem == ==Solution 1== We first count the number of ways to place <imath>4</imath> distinct people into <imath>6</imath> distinct chairs: <imath>6\cdot5\cdot4\cdot3 = 360</imath>. We now count how many favorable assignments there are. There are <imath>6</imath> ways to choose an adjacent pair of chairs for the two students. We can arrange the two students in those two chairs in <imath>2</imath> ways. After those two adjacent chairs are taken, there are <imath>4..." |
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== Problem == | == Problem == | ||
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? | |||
<imath>\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</imath> | |||
==Solution 1== | ==Solution 1== | ||
Revision as of 15:07, 6 November 2025
Problem
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?
Solution 1
We first count the number of ways to place 
distinct people into 
distinct chairs: 
.
We now count how many favorable assignments there are. There are ways to choose an adjacent pair of chairs for the two students. We can arrange the two students in those two chairs in 
ways.
After those two adjacent chairs are taken, there are chairs left, with 
adjacent pairs among them. We choose one of these pairs, arranging the teachers for 
ways.
There are 
favorable arrangements.
The probability is therefore 
~lprado
