Seating around a table • Probability two people sit together • Stepping around a circle
Circular Table Seating
When people sit around a circular table, rotations of the same arrangement are considered the same. The usual trick is to fix one person’s position and arrange everyone else relative to that person.
Example: 5 people around a table → fix one person, then arrange the other 4 in \(4! = 4x3x2x1 = 24\) ways.
Two People Sitting Together
Fix one person. The second person then has only 2 seats next to the first person (one on each side) out of the remaining seats.
Example: How many ways to seat two people next to each other at a 6 person circular table.
Seat the first person in a fixed location.
There are five seats left.
The other person could sit to the left or the right of the person.
Probability that A and B sit next to each other = \(\dfrac{2}{5}\).
Stepping Around a Circle
If you color every \(n\)th dot on a circle of \(m\) dots, you will eventually color every dot if and only if \(\gcd(n,m) = 1\).
Example: 12 dots, step size 5. Since \(\gcd(5,12)=1\) (no common factors other than 1), every dot will be colored.
The most common circular arrangement problem on the ACT.
Five friends are seated randomly around a circular table. What is the probability that two particular friends, Alex and Blake, sit next to each other?
Fix Alex’s position (this removes the circular symmetry).
There are 4 seats left for Blake.
Exactly 2 of those seats are next to Alex.
Probability = \(\dfrac{2}{4} = \dfrac{1}{2}\)
Answer: \(\dfrac{1}{2}\)
Kenji and Mary are members of a school committee. The 6 members of the committee will be seated randomly around a circular table. What is the probability that Kenji and Mary will NOT sit next to each other?
Fix Kenji’s position.
Mary has 5 possible seats left.
2 of those seats are next to Kenji, so the probability they are next to each other is \(\dfrac{2}{5}\).
Therefore the probability they are not next to each other is:
$$1 – \dfrac{2}{5} = \dfrac{3}{5}$$
Answer: \(\dfrac{3}{5}\)
Eight people are seated randomly around a circular table. What is the probability that two particular people, A and B, sit opposite each other?
Fix person A’s position (this removes the circular symmetry).
There are 7 seats left for B.
With 8 seats, exactly 1 seat is directly opposite A.
Probability = \(\dfrac{1}{7}\)
Answer: \(\dfrac{1}{7}\)
You will hit every dot if and only if the step size and the number of dots are relatively prime.
There are 12 equally spaced dots on a circle. Starting at one dot, you color every 5th dot and continue until you return to a colored dot. Will you color every dot?
Check whether \(\gcd(5,12) = 1\).
Yes, \(\gcd(5,12) = 1\), so you will color every dot.
Answer: Yes
There are 36 equally spaced dots marked on a circle. Starting at one dot, you color every \(k\)th dot and continue around the circle until you return to a previously colored dot. For which of the following values of \(k\) will every dot eventually be colored?
A. 4 B. 6 C. 8 D. 9 E. 11
You will color every dot if and only if \(\gcd(k,36)=1\).
36 = \(2^{2} \times 3^{2}\)
Answer: E. 11
There are 30 equally spaced dots marked on a circle. Starting at one dot, you color every \(k\)th dot and continue around the circle. For how many integers \(k\) from 1 to 29 will you eventually color every dot?
You color every dot if and only if \(\gcd(k,30)=1\).
30 = \(2 \times 3 \times 5\)
The positive integers less than 30 that are relatively prime to 30 are the numbers not divisible by 2, 3, or 5:
1, 7, 11, 13, 17, 19, 23, 29
There are 8 such numbers.
This is good practice, but harder than the questions on the ACT.
Answer: 8
Need help mastering circular arrangement problems for the ACT?