Find the number of ways of choosing three circles below, so that no two circles are next to each other.

[asy]

unitsize(1 cm);

int i;

for (i = 0; i <= 11; ++i) {

draw(Circle(2*dir(360/12*i),0.3));

label(string(i + 1), 2*dir(90 - 360/12*i));

}

[/asy]

ghistory Oct 13, 2024

#1**0 **

To find the number of ways to choose 3 circles from 12 arranged in a circle such that no two circles are adjacent, we can use a combinatorial approach.

First, since the circles are arranged in a circle, we'll convert the problem into a linear arrangement. We can fix one circle as a reference point, which effectively reduces our problem to a line of 11 circles (with the circle we fixed being considered as already chosen).

Once we fix a circle, we need to select 2 more circles from the remaining 11 while ensuring that no two chosen circles are adjacent. To achieve this, we can represent the chosen circles and the unchosen circles as follows:

If we choose two additional circles, there will be at least one unchosen circle between any two chosen circles. Therefore, we can denote chosen circles as 'C' and unchosen circles as 'U'. So, if we denote our arrangement with chosen circles and a required space, it will look something like this:

- C - U - C - U - C

This configuration illustrates that for each chosen circle, 1 unchosen circle must be placed between them, leading us to count the total number of positions left for unchosen circles after accounting for the chosen ones.

### Create the remaining sequence for unchosen circles:

Given:

- We have chosen 3 'C's, thus requiring 2 'U's to separate them.

- This means we use up 3 + 2 = 5 spots out of the 11 available, leaving us with 6 unchosen circles to place.

### Transform into a stars and bars problem:

Now we need to distribute the remaining unchosen circles among available spots. We can think of placing the 6 remaining unchosen circles in gaps created by the two boundaries (left and right of 'C's):

If we denote the gaps as follows (where '|' represents gaps):

- | U | U | C | U | C | U | C | U | |

We have potential gaps before the first 'C', between the 'C's, and after the last 'C'. In our case, we have fixed the first circle, which means we currently have two gaps to the left and one gap to the right (which connects back to the fixed circle in circular arrangement).

- Let G - denotes the gap available, thus a sequence could be modeled as:

- G (pre gap), C, U, C, U, C (for the already occupied spaces) plus additional G's around.

### Fill spaces with the left-over circles:

Thus, after fixing the first chosen circle, thus we have 8 available spots (in gaps) to place our remaining circles, as constructed above, calculating the arrangement with stars and bars theorem.

Using the formula for distributing \( n \) indistinguishable objects (stars) into \( k \) distinguishable boxes (spaces):

\[ \text{total ways} = \binom{n+k-1}{k-1} \]

where \( n = 6 \) (remaining unchosen circles) and \( k = 3 \) (gaps). Substitute this into the equation:

\[

\text{total ways} = \binom{6 + 3 - 1}{3 - 1} = \binom{8}{2} = 28

\]

### Final Answer:

Thus, the total number of ways to choose the circles such that no two are adjacent is \( \boxed{220} \) ways.

Akhain1 Oct 13, 2024