Balancing Balls
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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:
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
(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
asked Dec 8 '18 at 6:06
Dr XorileDr Xorile
12k22566
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add a comment |
$begingroup$
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:
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
(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
asked Dec 8 '18 at 6:06
Dr XorileDr Xorile
12k22566
$endgroup$
1
$begingroup$
Are you asking about the number of marbles, or the combinations of the number of marbles?
$endgroup$
– Wais Kamal
Dec 8 '18 at 8:14
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do we need physic tag since it is about the balancing the disc?
$endgroup$
– Oray
Dec 8 '18 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 '18 at 17:02
add a comment |
$begingroup$
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:
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
(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
asked Dec 8 '18 at 6:06
Dr XorileDr Xorile
12k22566
$endgroup$
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:
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
(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
combinatorics physics
asked Dec 8 '18 at 6:06
Dr XorileDr Xorile
12k22566
asked Dec 8 '18 at 6:06
Dr XorileDr Xorile
12k22566
edited Dec 8 '18 at 17:22
Dr Xorile
asked Dec 8 '18 at 6:06
Dr XorileDr Xorile
12k22566
asked Dec 8 '18 at 6:06
Dr XorileDr Xorile
12k22566
asked Dec 8 '18 at 6:06
Dr XorileDr Xorile
12k22566
12k22566
1
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Are you asking about the number of marbles, or the combinations of the number of marbles?
$endgroup$
– Wais Kamal
Dec 8 '18 at 8:14
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do we need physic tag since it is about the balancing the disc?
$endgroup$
– Oray
Dec 8 '18 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 '18 at 17:02
add a comment |
1
$begingroup$
Are you asking about the number of marbles, or the combinations of the number of marbles?
$endgroup$
– Wais Kamal
Dec 8 '18 at 8:14
$begingroup$
do we need physic tag since it is about the balancing the disc?
$endgroup$
– Oray
Dec 8 '18 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 '18 at 17:02
1
1
$begingroup$
Are you asking about the number of marbles, or the combinations of the number of marbles?
$endgroup$
– Wais Kamal
Dec 8 '18 at 8:14
$begingroup$
Are you asking about the number of marbles, or the combinations of the number of marbles?
$endgroup$
– Wais Kamal
Dec 8 '18 at 8:14
$begingroup$
do we need physic tag since it is about the balancing the disc?
$endgroup$
– Oray
Dec 8 '18 at 14:42
$begingroup$
do we need physic tag since it is about the balancing the disc?
$endgroup$
– Oray
Dec 8 '18 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 '18 at 17:02
$begingroup$
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 '18 at 17:02
add a comment |
4 Answers
4
active
oldest
votes
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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:
answered Dec 8 '18 at 9:02
Gareth McCaughan♦Gareth McCaughan
61.6k3152238
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what about oxoxoxoxoxoo? and some others.
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– JonMark Perry
Dec 8 '18 at 10:11
2
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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".
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– Gareth McCaughan♦
Dec 8 '18 at 10:33
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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.
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– BmyGuest
Dec 10 '18 at 10:19
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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 '18 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.)
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– Gareth McCaughan♦
Dec 10 '18 at 10:56
add a comment |
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Another answer has missed one permutation of
9 balls
Note that each solution has its complement too.
answered Dec 8 '18 at 8:37
Weather VaneWeather Vane
1,36419
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add a comment |
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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:
Using $6$ marbles:
answered Dec 8 '18 at 8:38
JonMark PerryJonMark Perry
19k63891
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add a comment |
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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.
answered Dec 8 '18 at 8:12
Wais KamalWais Kamal
412111
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rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
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– jafe
Dec 8 '18 at 8:37
1
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@jafe I was posting that as you commented.
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– Weather Vane
Dec 8 '18 at 8:38
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@jafe thanks for pointing it out. I edited my answer.
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– Wais Kamal
Dec 8 '18 at 11:11
1
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But the OP doesn't count rotations or reflections.
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– amI
Dec 8 '18 at 13:05
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The OP didn't say something about rotations or reflections.
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– Wais Kamal
Dec 8 '18 at 13:18
|
show 1 more comment
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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:
answered Dec 8 '18 at 9:02
Gareth McCaughan♦Gareth McCaughan
61.6k3152238
$endgroup$
$begingroup$
what about oxoxoxoxoxoo? and some others.
$endgroup$
– JonMark Perry
Dec 8 '18 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 '18 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 '18 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 '18 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 '18 at 10:56
add a comment |
$begingroup$
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:
answered Dec 8 '18 at 9:02
Gareth McCaughan♦Gareth McCaughan
61.6k3152238
$endgroup$
$begingroup$
what about oxoxoxoxoxoo? and some others.
$endgroup$
– JonMark Perry
Dec 8 '18 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 '18 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 '18 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 '18 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 '18 at 10:56
add a comment |
$begingroup$
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:
answered Dec 8 '18 at 9:02
Gareth McCaughan♦Gareth McCaughan
61.6k3152238
$endgroup$
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:
answered Dec 8 '18 at 9:02
Gareth McCaughan♦Gareth McCaughan
61.6k3152238
edited Dec 10 '18 at 10:57
answered Dec 8 '18 at 9:02
Gareth McCaughan♦Gareth McCaughan
61.6k3152238
answered Dec 8 '18 at 9:02
Gareth McCaughan♦Gareth McCaughan
61.6k3152238
answered Dec 8 '18 at 9:02
Gareth McCaughan♦Gareth McCaughan
61.6k3152238
61.6k3152238
$begingroup$
what about oxoxoxoxoxoo? and some others.
$endgroup$
– JonMark Perry
Dec 8 '18 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 '18 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 '18 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 '18 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 '18 at 10:56
add a comment |
$begingroup$
what about oxoxoxoxoxoo? and some others.
$endgroup$
– JonMark Perry
Dec 8 '18 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 '18 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 '18 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 '18 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 '18 at 10:56
$begingroup$
what about oxoxoxoxoxoo? and some others.
$endgroup$
– JonMark Perry
Dec 8 '18 at 10:11
$begingroup$
what about oxoxoxoxoxoo? and some others.
$endgroup$
– JonMark Perry
Dec 8 '18 at 10:11
2
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 '18 at 10:33
$begingroup$
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 '18 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 '18 at 10:19
$begingroup$
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 '18 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 '18 at 10:55
$begingroup$
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 '18 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 '18 at 10:56
$begingroup$
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 '18 at 10:56
add a comment |
$begingroup$
Another answer has missed one permutation of
9 balls
Note that each solution has its complement too.
answered Dec 8 '18 at 8:37
Weather VaneWeather Vane
1,36419
$endgroup$
add a comment |
$begingroup$
Another answer has missed one permutation of
9 balls
Note that each solution has its complement too.
answered Dec 8 '18 at 8:37
Weather VaneWeather Vane
1,36419
$endgroup$
add a comment |
$begingroup$
Another answer has missed one permutation of
9 balls
Note that each solution has its complement too.
answered Dec 8 '18 at 8:37
Weather VaneWeather Vane
1,36419
$endgroup$
Another answer has missed one permutation of
9 balls
Note that each solution has its complement too.
answered Dec 8 '18 at 8:37
Weather VaneWeather Vane
1,36419
answered Dec 8 '18 at 8:37
Weather VaneWeather Vane
1,36419
answered Dec 8 '18 at 8:37
Weather VaneWeather Vane
1,36419
answered Dec 8 '18 at 8:37
Weather VaneWeather Vane
1,36419
1,36419
add a comment |
add a comment |
$begingroup$
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:
Using $6$ marbles:
answered Dec 8 '18 at 8:38
JonMark PerryJonMark Perry
19k63891
$endgroup$
add a comment |
$begingroup$
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:
Using $6$ marbles:
answered Dec 8 '18 at 8:38
JonMark PerryJonMark Perry
19k63891
$endgroup$
add a comment |
$begingroup$
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:
Using $6$ marbles:
answered Dec 8 '18 at 8:38
JonMark PerryJonMark Perry
19k63891
$endgroup$
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:
Using $6$ marbles:
answered Dec 8 '18 at 8:38
JonMark PerryJonMark Perry
19k63891
edited Dec 14 '18 at 8:40
answered Dec 8 '18 at 8:38
JonMark PerryJonMark Perry
19k63891
answered Dec 8 '18 at 8:38
JonMark PerryJonMark Perry
19k63891
answered Dec 8 '18 at 8:38
JonMark PerryJonMark Perry
19k63891
19k63891
add a comment |
add a comment |
$begingroup$
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.
answered Dec 8 '18 at 8:12
Wais KamalWais Kamal
412111
$endgroup$
$begingroup$
rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
$endgroup$
– jafe
Dec 8 '18 at 8:37
1
$begingroup$
@jafe I was posting that as you commented.
$endgroup$
– Weather Vane
Dec 8 '18 at 8:38
$begingroup$
@jafe thanks for pointing it out. I edited my answer.
$endgroup$
– Wais Kamal
Dec 8 '18 at 11:11
1
$begingroup$
But the OP doesn't count rotations or reflections.
$endgroup$
– amI
Dec 8 '18 at 13:05
$begingroup$
The OP didn't say something about rotations or reflections.
$endgroup$
– Wais Kamal
Dec 8 '18 at 13:18
|
show 1 more comment
$begingroup$
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.
answered Dec 8 '18 at 8:12
Wais KamalWais Kamal
412111
$endgroup$
$begingroup$
rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
$endgroup$
– jafe
Dec 8 '18 at 8:37
1
$begingroup$
@jafe I was posting that as you commented.
$endgroup$
– Weather Vane
Dec 8 '18 at 8:38
$begingroup$
@jafe thanks for pointing it out. I edited my answer.
$endgroup$
– Wais Kamal
Dec 8 '18 at 11:11
1
$begingroup$
But the OP doesn't count rotations or reflections.
$endgroup$
– amI
Dec 8 '18 at 13:05
$begingroup$
The OP didn't say something about rotations or reflections.
$endgroup$
– Wais Kamal
Dec 8 '18 at 13:18
|
show 1 more comment
$begingroup$
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.
answered Dec 8 '18 at 8:12
Wais KamalWais Kamal
412111
$endgroup$
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.
answered Dec 8 '18 at 8:12
Wais KamalWais Kamal
412111
edited Dec 8 '18 at 11:26
answered Dec 8 '18 at 8:12
Wais KamalWais Kamal
412111
answered Dec 8 '18 at 8:12
Wais KamalWais Kamal
412111
answered Dec 8 '18 at 8:12
Wais KamalWais Kamal
412111
412111
$begingroup$
rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
$endgroup$
– jafe
Dec 8 '18 at 8:37
1
$begingroup$
@jafe I was posting that as you commented.
$endgroup$
– Weather Vane
Dec 8 '18 at 8:38
$begingroup$
@jafe thanks for pointing it out. I edited my answer.
$endgroup$
– Wais Kamal
Dec 8 '18 at 11:11
1
$begingroup$
But the OP doesn't count rotations or reflections.
$endgroup$
– amI
Dec 8 '18 at 13:05
$begingroup$
The OP didn't say something about rotations or reflections.
$endgroup$
– Wais Kamal
Dec 8 '18 at 13:18
|
show 1 more comment
$begingroup$
rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
$endgroup$
– jafe
Dec 8 '18 at 8:37
1
$begingroup$
@jafe I was posting that as you commented.
$endgroup$
– Weather Vane
Dec 8 '18 at 8:38
$begingroup$
@jafe thanks for pointing it out. I edited my answer.
$endgroup$
– Wais Kamal
Dec 8 '18 at 11:11
1
$begingroup$
But the OP doesn't count rotations or reflections.
$endgroup$
– amI
Dec 8 '18 at 13:05
$begingroup$
The OP didn't say something about rotations or reflections.
$endgroup$
– Wais Kamal
Dec 8 '18 at 13:18
$begingroup$
rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
$endgroup$
– jafe
Dec 8 '18 at 8:37
$begingroup$
rot13[Vs guerr jbexf, fubhyqa'g avar jbex nf jryy?]
$endgroup$
– jafe
Dec 8 '18 at 8:37
1
1
$begingroup$
@jafe I was posting that as you commented.
$endgroup$
– Weather Vane
Dec 8 '18 at 8:38
$begingroup$
@jafe I was posting that as you commented.
$endgroup$
– Weather Vane
Dec 8 '18 at 8:38
$begingroup$
@jafe thanks for pointing it out. I edited my answer.
$endgroup$
– Wais Kamal
Dec 8 '18 at 11:11
$begingroup$
@jafe thanks for pointing it out. I edited my answer.
$endgroup$
– Wais Kamal
Dec 8 '18 at 11:11
1
1
$begingroup$
But the OP doesn't count rotations or reflections.
$endgroup$
– amI
Dec 8 '18 at 13:05
$begingroup$
But the OP doesn't count rotations or reflections.
$endgroup$
– amI
Dec 8 '18 at 13:05
$begingroup$
The OP didn't say something about rotations or reflections.
$endgroup$
– Wais Kamal
Dec 8 '18 at 13:18
$begingroup$
The OP didn't say something about rotations or reflections.
$endgroup$
– Wais Kamal
Dec 8 '18 at 13:18
|
show 1 more comment
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1
$begingroup$
Are you asking about the number of marbles, or the combinations of the number of marbles?
$endgroup$
– Wais Kamal
Dec 8 '18 at 8:14
$begingroup$
do we need physic tag since it is about the balancing the disc?
$endgroup$
– Oray
Dec 8 '18 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 '18 at 17:02