Balancing Balls











up vote
14
down vote

favorite
3












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











share|improve this question




















  • 1




    Are you asking about the number of marbles, or the combinations of the number of marbles?
    – Wais Kamal
    Dec 8 at 8:14










  • do we need physic tag since it is about the balancing the disc?
    – Oray
    Dec 8 at 14:42










  • 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.
    – Dr Xorile
    Dec 8 at 17:02















up vote
14
down vote

favorite
3












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











share|improve this question




















  • 1




    Are you asking about the number of marbles, or the combinations of the number of marbles?
    – Wais Kamal
    Dec 8 at 8:14










  • do we need physic tag since it is about the balancing the disc?
    – Oray
    Dec 8 at 14:42










  • 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.
    – Dr Xorile
    Dec 8 at 17:02













up vote
14
down vote

favorite
3









up vote
14
down vote

favorite
3






3





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











share|improve this question















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








combinatorics physics






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 8 at 17:22

























asked Dec 8 at 6:06









Dr Xorile

11.3k12363




11.3k12363








  • 1




    Are you asking about the number of marbles, or the combinations of the number of marbles?
    – Wais Kamal
    Dec 8 at 8:14










  • do we need physic tag since it is about the balancing the disc?
    – Oray
    Dec 8 at 14:42










  • 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.
    – Dr Xorile
    Dec 8 at 17:02














  • 1




    Are you asking about the number of marbles, or the combinations of the number of marbles?
    – Wais Kamal
    Dec 8 at 8:14










  • do we need physic tag since it is about the balancing the disc?
    – Oray
    Dec 8 at 14:42










  • 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.
    – Dr Xorile
    Dec 8 at 17:02








1




1




Are you asking about the number of marbles, or the combinations of the number of marbles?
– Wais Kamal
Dec 8 at 8:14




Are you asking about the number of marbles, or the combinations of the number of marbles?
– Wais Kamal
Dec 8 at 8:14












do we need physic tag since it is about the balancing the disc?
– Oray
Dec 8 at 14:42




do we need physic tag since it is about the balancing the disc?
– Oray
Dec 8 at 14:42












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.
– Dr Xorile
Dec 8 at 17:02




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.
– Dr Xorile
Dec 8 at 17:02










4 Answers
4






active

oldest

votes

















up vote
13
down vote



accepted










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:




Solution







share|improve this answer























  • what about oxoxoxoxoxoo? and some others.
    – JonMark Perry
    Dec 8 at 10:11






  • 2




    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".
    – Gareth McCaughan
    Dec 8 at 10:33










  • 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.
    – BmyGuest
    Dec 10 at 10:19










  • The reason why you don't get those comments is that they were stupidly wrong. I will remove them. Thanks for adding the pictures!
    – Gareth McCaughan
    Dec 10 at 10:55










  • Sorry, to be more precise: one of them does have mirror symmetry, the other doesn't. (It has a different sort of symmetry, though.)
    – Gareth McCaughan
    Dec 10 at 10:56


















up vote
3
down vote













Another answer has missed one permutation of




9 balls
enter image description here




Note that each solution has its complement too.






share|improve this answer





















  • Added to my answer, thanks!
    – Wais Kamal
    Dec 8 at 11:30


















up vote
1
down vote













The answer is:




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




because




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:
circles6_1circles6_2circles6_3







share|improve this answer






























    up vote
    0
    down vote













    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.







    share|improve this answer























    • rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
      – jafe
      Dec 8 at 8:37






    • 1




      @jafe I was posting that as you commented.
      – Weather Vane
      Dec 8 at 8:38










    • @jafe thanks for pointing it out. I edited my answer.
      – Wais Kamal
      Dec 8 at 11:11






    • 1




      But the OP doesn't count rotations or reflections.
      – amI
      Dec 8 at 13:05










    • The OP didn't say something about rotations or reflections.
      – Wais Kamal
      Dec 8 at 13:18











    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "559"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    convertImagesToLinks: false,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: null,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f76191%2fbalancing-balls%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    13
    down vote



    accepted










    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:




    Solution







    share|improve this answer























    • what about oxoxoxoxoxoo? and some others.
      – JonMark Perry
      Dec 8 at 10:11






    • 2




      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".
      – Gareth McCaughan
      Dec 8 at 10:33










    • 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.
      – BmyGuest
      Dec 10 at 10:19










    • The reason why you don't get those comments is that they were stupidly wrong. I will remove them. Thanks for adding the pictures!
      – Gareth McCaughan
      Dec 10 at 10:55










    • Sorry, to be more precise: one of them does have mirror symmetry, the other doesn't. (It has a different sort of symmetry, though.)
      – Gareth McCaughan
      Dec 10 at 10:56















    up vote
    13
    down vote



    accepted










    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:




    Solution







    share|improve this answer























    • what about oxoxoxoxoxoo? and some others.
      – JonMark Perry
      Dec 8 at 10:11






    • 2




      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".
      – Gareth McCaughan
      Dec 8 at 10:33










    • 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.
      – BmyGuest
      Dec 10 at 10:19










    • The reason why you don't get those comments is that they were stupidly wrong. I will remove them. Thanks for adding the pictures!
      – Gareth McCaughan
      Dec 10 at 10:55










    • Sorry, to be more precise: one of them does have mirror symmetry, the other doesn't. (It has a different sort of symmetry, though.)
      – Gareth McCaughan
      Dec 10 at 10:56













    up vote
    13
    down vote



    accepted







    up vote
    13
    down vote



    accepted






    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:




    Solution







    share|improve this answer














    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:




    Solution








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Dec 10 at 10:57

























    answered Dec 8 at 9:02









    Gareth McCaughan

    60.1k3150232




    60.1k3150232












    • what about oxoxoxoxoxoo? and some others.
      – JonMark Perry
      Dec 8 at 10:11






    • 2




      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".
      – Gareth McCaughan
      Dec 8 at 10:33










    • 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.
      – BmyGuest
      Dec 10 at 10:19










    • The reason why you don't get those comments is that they were stupidly wrong. I will remove them. Thanks for adding the pictures!
      – Gareth McCaughan
      Dec 10 at 10:55










    • Sorry, to be more precise: one of them does have mirror symmetry, the other doesn't. (It has a different sort of symmetry, though.)
      – Gareth McCaughan
      Dec 10 at 10:56


















    • what about oxoxoxoxoxoo? and some others.
      – JonMark Perry
      Dec 8 at 10:11






    • 2




      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".
      – Gareth McCaughan
      Dec 8 at 10:33










    • 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.
      – BmyGuest
      Dec 10 at 10:19










    • The reason why you don't get those comments is that they were stupidly wrong. I will remove them. Thanks for adding the pictures!
      – Gareth McCaughan
      Dec 10 at 10:55










    • Sorry, to be more precise: one of them does have mirror symmetry, the other doesn't. (It has a different sort of symmetry, though.)
      – Gareth McCaughan
      Dec 10 at 10:56
















    what about oxoxoxoxoxoo? and some others.
    – JonMark Perry
    Dec 8 at 10:11




    what about oxoxoxoxoxoo? and some others.
    – JonMark Perry
    Dec 8 at 10:11




    2




    2




    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".
    – Gareth McCaughan
    Dec 8 at 10:33




    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".
    – Gareth McCaughan
    Dec 8 at 10:33












    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.
    – BmyGuest
    Dec 10 at 10:19




    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.
    – BmyGuest
    Dec 10 at 10:19












    The reason why you don't get those comments is that they were stupidly wrong. I will remove them. Thanks for adding the pictures!
    – Gareth McCaughan
    Dec 10 at 10:55




    The reason why you don't get those comments is that they were stupidly wrong. I will remove them. Thanks for adding the pictures!
    – Gareth McCaughan
    Dec 10 at 10:55












    Sorry, to be more precise: one of them does have mirror symmetry, the other doesn't. (It has a different sort of symmetry, though.)
    – Gareth McCaughan
    Dec 10 at 10:56




    Sorry, to be more precise: one of them does have mirror symmetry, the other doesn't. (It has a different sort of symmetry, though.)
    – Gareth McCaughan
    Dec 10 at 10:56










    up vote
    3
    down vote













    Another answer has missed one permutation of




    9 balls
    enter image description here




    Note that each solution has its complement too.






    share|improve this answer





















    • Added to my answer, thanks!
      – Wais Kamal
      Dec 8 at 11:30















    up vote
    3
    down vote













    Another answer has missed one permutation of




    9 balls
    enter image description here




    Note that each solution has its complement too.






    share|improve this answer





















    • Added to my answer, thanks!
      – Wais Kamal
      Dec 8 at 11:30













    up vote
    3
    down vote










    up vote
    3
    down vote









    Another answer has missed one permutation of




    9 balls
    enter image description here




    Note that each solution has its complement too.






    share|improve this answer












    Another answer has missed one permutation of




    9 balls
    enter image description here




    Note that each solution has its complement too.







    share|improve this answer












    share|improve this answer



    share|improve this answer










    answered Dec 8 at 8:37









    Weather Vane

    1,01719




    1,01719












    • Added to my answer, thanks!
      – Wais Kamal
      Dec 8 at 11:30


















    • Added to my answer, thanks!
      – Wais Kamal
      Dec 8 at 11:30
















    Added to my answer, thanks!
    – Wais Kamal
    Dec 8 at 11:30




    Added to my answer, thanks!
    – Wais Kamal
    Dec 8 at 11:30










    up vote
    1
    down vote













    The answer is:




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




    because




    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:
    circles6_1circles6_2circles6_3







    share|improve this answer



























      up vote
      1
      down vote













      The answer is:




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




      because




      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:
      circles6_1circles6_2circles6_3







      share|improve this answer

























        up vote
        1
        down vote










        up vote
        1
        down vote









        The answer is:




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




        because




        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:
        circles6_1circles6_2circles6_3







        share|improve this answer














        The answer is:




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




        because




        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:
        circles6_1circles6_2circles6_3








        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited yesterday

























        answered Dec 8 at 8:38









        JonMark Perry

        17k63281




        17k63281






















            up vote
            0
            down vote













            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.







            share|improve this answer























            • rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
              – jafe
              Dec 8 at 8:37






            • 1




              @jafe I was posting that as you commented.
              – Weather Vane
              Dec 8 at 8:38










            • @jafe thanks for pointing it out. I edited my answer.
              – Wais Kamal
              Dec 8 at 11:11






            • 1




              But the OP doesn't count rotations or reflections.
              – amI
              Dec 8 at 13:05










            • The OP didn't say something about rotations or reflections.
              – Wais Kamal
              Dec 8 at 13:18















            up vote
            0
            down vote













            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.







            share|improve this answer























            • rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
              – jafe
              Dec 8 at 8:37






            • 1




              @jafe I was posting that as you commented.
              – Weather Vane
              Dec 8 at 8:38










            • @jafe thanks for pointing it out. I edited my answer.
              – Wais Kamal
              Dec 8 at 11:11






            • 1




              But the OP doesn't count rotations or reflections.
              – amI
              Dec 8 at 13:05










            • The OP didn't say something about rotations or reflections.
              – Wais Kamal
              Dec 8 at 13:18













            up vote
            0
            down vote










            up vote
            0
            down vote









            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.







            share|improve this answer














            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.








            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited Dec 8 at 11:26

























            answered Dec 8 at 8:12









            Wais Kamal

            412111




            412111












            • rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
              – jafe
              Dec 8 at 8:37






            • 1




              @jafe I was posting that as you commented.
              – Weather Vane
              Dec 8 at 8:38










            • @jafe thanks for pointing it out. I edited my answer.
              – Wais Kamal
              Dec 8 at 11:11






            • 1




              But the OP doesn't count rotations or reflections.
              – amI
              Dec 8 at 13:05










            • The OP didn't say something about rotations or reflections.
              – Wais Kamal
              Dec 8 at 13:18


















            • rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
              – jafe
              Dec 8 at 8:37






            • 1




              @jafe I was posting that as you commented.
              – Weather Vane
              Dec 8 at 8:38










            • @jafe thanks for pointing it out. I edited my answer.
              – Wais Kamal
              Dec 8 at 11:11






            • 1




              But the OP doesn't count rotations or reflections.
              – amI
              Dec 8 at 13:05










            • The OP didn't say something about rotations or reflections.
              – Wais Kamal
              Dec 8 at 13:18
















            rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
            – jafe
            Dec 8 at 8:37




            rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
            – jafe
            Dec 8 at 8:37




            1




            1




            @jafe I was posting that as you commented.
            – Weather Vane
            Dec 8 at 8:38




            @jafe I was posting that as you commented.
            – Weather Vane
            Dec 8 at 8:38












            @jafe thanks for pointing it out. I edited my answer.
            – Wais Kamal
            Dec 8 at 11:11




            @jafe thanks for pointing it out. I edited my answer.
            – Wais Kamal
            Dec 8 at 11:11




            1




            1




            But the OP doesn't count rotations or reflections.
            – amI
            Dec 8 at 13:05




            But the OP doesn't count rotations or reflections.
            – amI
            Dec 8 at 13:05












            The OP didn't say something about rotations or reflections.
            – Wais Kamal
            Dec 8 at 13:18




            The OP didn't say something about rotations or reflections.
            – Wais Kamal
            Dec 8 at 13:18


















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Puzzling Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f76191%2fbalancing-balls%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            Сан-Квентин

            Алькесар

            Josef Freinademetz