I have a disk with 6 equally spaced dents around the edge. The disk balances on the center point. I want to place marbles around the edge so that it stays balanced. There are four ways that this can be done:

6 balance

One with two balls, one with three, one with four, and one with six. It can't be balanced with five or one.

Now the question is:

How many ways (all combinations excluding rotations and reflections) are there to balance the disk if it has 12 dents
12 balance
(Imagine that my drawing skills were better and the 12 dents are all equally spaced!)

Now that there's a valid answer, I'm going to add a source with the inspiration and an interesting fact about it:

As some of the answerers noted, each answer has a complement. If it can be balanced in one configuration, then you can swap every marble for a hole and every hole for a marble and it will still be balanced.
It turns out that with $n$ holes and $k$ marbles it can be balanced if and only if both $k$ and $n-k$ can be written as the sum of prime factors of $n$.
Here's a link to the numberphile video that inspired this: Numberphile video

  • 1
    $\begingroup$ Are you asking about the number of marbles, or the combinations of the number of marbles? $\endgroup$
    – Wais Kamal
    Dec 8, 2018 at 8:14
  • $\begingroup$ do we need physic tag since it is about the balancing the disc? $\endgroup$
    – Oray
    Dec 8, 2018 at 14:42
  • $\begingroup$ Combinations is what I'm looking for. I'll add a physics tag if that's the consensus. Mathematically, the sum of the points represented by the marbles needs to be 0. $\endgroup$
    – Dr Xorile
    Dec 8, 2018 at 17:02

4 Answers 4


There are more possibilities than one expects because

perhaps surprisingly, neither the number of balls nor the number of holes need be a factor of 12! (You can put together a solution with 2 and a solution with 3 to get one with 5, for instance.)

Here is the full list, after removing symmetries; I have marked some of the more surprising ones:

............ oo..oo..oo.. o.o.o.o.o.o. o.....o..... oooo..oooo.. o..o..o..o.. oo....oo.... ooo..oo.oo.. ** (seven balls!) ooo.ooo.ooo. ooooo.ooooo. o...o...o... o.o...o.o... ooo.o.ooo.o. oo.oo.oo.oo. oooooooooooo oo..o..oo... ** (five balls!) ooo...ooo... oo.o..oo.o.. * (no mirror symmetry)

So there are

18 ways to do it. This includes the "empty" solution; OP implies that this one isn't acceptable, so the answer OP is looking for is 17.

A visual representation:


  • $\begingroup$ what about oxoxoxoxoxoo? and some others. $\endgroup$
    – JMP
    Dec 8, 2018 at 10:11
  • 2
    $\begingroup$ The one you give doesn't balance. (Replace the last "o" with an "x" and it obviously does; therefore, as it stands it doesn't.) I can't comment on an unspecified "some others". $\endgroup$
    – Gareth McCaughan
    Dec 8, 2018 at 10:33
  • $\begingroup$ I don't get your "no-mirror symmetry" comments. Both those configurations clearly have mirror symmetry ?? ooo.o.ooo.o. even has two mirror-axes. (See my now edited in picture.) Otherwise great solution. $\endgroup$
    – BmyGuest
    Dec 10, 2018 at 10:19
  • $\begingroup$ The reason why you don't get those comments is that they were stupidly wrong. I will remove them. Thanks for adding the pictures! $\endgroup$
    – Gareth McCaughan
    Dec 10, 2018 at 10:55
  • $\begingroup$ Sorry, to be more precise: one of them does have mirror symmetry, the other doesn't. (It has a different sort of symmetry, though.) $\endgroup$
    – Gareth McCaughan
    Dec 10, 2018 at 10:56

Another answer has missed one permutation of

9 balls
enter image description here

Note that each solution has its complement too.


The answer is:

$15$ (counting $0$ marbles as an option)


For a board to balance, the resultant force must be zero.

$k$ must therefore be a divisor of $12$ greater than $1$ to achieve a balance, hence $2,3,4,6,12$, but leaving these as holes also balances the board, and so we also have $10,9,8,6,0$, which makes for $9$ cases in total. There are $3$ versions with $4$ marbles (and therefore also with $8$ marbles), and $3$ versions with $6$ marbles, giving $15$ in total.

Using $4$ marbles:
circles4_1circles4_2enter image description here

Using $6$ marbles:


I will start solving the questions by mentioning the two required solution techniques.

Technique 1:

The marbles can be arranged such that every pair of marbles are facing each other (used with an even number of marbles). The total number of ways is given by $^{N/2}C_{n/2}$, where $N$ is the total number of dents, and $n$ is the number of marbles to be used.

Technique 2:

The marbles can be arranged such that they are separated by an equal number of dents. (used with an odd number of marbles).

Using these two techniques, the number of possible combinations for each number of marbles can be calculated.

2 marbles:

They can be arranged in $^6C_1 = 6$ ways.

3 marbles:

They can be arranged in $4$ ways.

4 marbles:

They can be arranged in $^6C_2 = 15$ ways.

6 marbles:

They can be arranged in $^6C_3 = 20$ ways.

8 marbles:

They can be arranged in $^6C_4 = 15$ ways.

9 marbles:

They can be arranged in 2 ways.

10 marbles:

They can be arranges in $^6C_5 = 6$ ways.

12 marbles:

There is only one way of arranging them ($^6C_1 = 1$).

One more thing...

The disk will also balance without any marbles.

Hence, the total number of arrangements that can be used to balance the disk is

$6 + 4 + 15 + 20 + 15 + 2 + 6 + 1 + 1 = 70$ ways.

  • $\begingroup$ rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?] $\endgroup$
    – Jafe
    Dec 8, 2018 at 8:37
  • 1
    $\begingroup$ @jafe I was posting that as you commented. $\endgroup$ Dec 8, 2018 at 8:38
  • $\begingroup$ @jafe thanks for pointing it out. I edited my answer. $\endgroup$
    – Wais Kamal
    Dec 8, 2018 at 11:11
  • 1
    $\begingroup$ But the OP doesn't count rotations or reflections. $\endgroup$
    – amI
    Dec 8, 2018 at 13:05
  • $\begingroup$ The OP didn't say something about rotations or reflections. $\endgroup$
    – Wais Kamal
    Dec 8, 2018 at 13:18

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