Dice problem

[hgupta]

There is a six faced die.Numbering should be done from 1 to 6.How many ways the numbering can be done , with a condition that 1 and 6, 2 and 5 , 3 and 4 comes on the opposite sides of the die.

is the answer 64??

[ajb]

Isn't it the number of permutations of 6 objects?

[hgupta]

sorry about my previous answer. is it 48??

 

my bad, thought that a dice has 8 faces!!

 

6 faces, so 6 options for 1 and 6 on opposite sides, consecutively 4 options for 2 and 5, and so 2 options for 3 and 4.

 

am i right??

[Aeternus]

Are you allowed to assume that each face is unique somehow or do you have to take into account that several different setups will work out the same once you rotate them?

[ajb]

Isn't the question how may ways can you arrange 6 objects?

 

suppose we have no numbers on the dice.

 

pick a number. how many sides can we attatch this number? The answer is 6.

 

pick another number. This can now only be placed in one of the remaining 5 sides.

 

pick another number. we now only have 4 sides to place it.

 

etc...

 

so we have the number of possible ways os assigning {1,2,3,4,5,6} to the six sides of a dice is

 

6.5.4.3.2.1 = 6! = 720

[Aeternus]

ajb, no, first of all he is restricting the number of choices, because certain numbers must be opposite each other, so the placement of one number immediately determines the placement of the other. Second of all, depending on the way the original question was intended, you can produce distinct dice "nets" that would appear as seperate possibilities but are essentially the same dice.

 

For Instance take -

 

--1

--2

3-6-4

--5

 

and

 

--6

--2

4-1 3

--5

 

They would seem to be different possibilities but in fact are just rotations of each other.

[ajb]

Taking into account the rotations you have 30 different die.

 

You can see you have 13 axis to rotate about, which gives you 24 turnings. Hence there are 720/24 = 30 different die.

 

Have a look at this website

 

http://www.mathematische-basteleien.de/dice.htm

[Aeternus]

You still aren't taking into account the fact that the numbers must be opposite each other.

 

Given that you are taking away the possibility of those opposite numbers, you have the 30 die you mention but you must divide that by 15 giving you two. The basic idea here is that you end up always having a vertex on the cube with 1, 2 and 3 meeting and the only option left to you is whether or not the numbers go around that vertex clockwise or anticlockwise as all the other ways of changing the numbers are just rotations of one of the others.

 

You can also think of it like this. Normally, as you said, you would do 6 x 5 x 4 x 3 x 2 x 1, but then as hgupta said, you would need to do 6 x 4 x 2 in this case as the first number's choice removes the 5 possibilities for the second, then the thirds removes the 3 for the 4th (as they HAVE to be opposite). This would leave us with 720/15 = 48 as hgupta said. However, with the rotations, as you said the number has to be divided by 24, so 48/24 = 2.

 

I could be wrong, I am rather tired and my head isn't necessarily in the right place but meh.

[ajb]

Ok, sorry, I did not look at the condition at all!

 

From my 30, you could then impose the condition and see what you get.

[Rajdilawar S]

Hi,

I think i got the soln for ur problem.Its a simple permutation problem

3!*3!*2!= 72.

We have three pairs i.e. can be arrange in 3 factorial was. And again factorial appears due to three pairs which can arrange in 3! ways.

And 2! for a pair to arrange.

[Jagged]

There is one unspecified thing about this problem. As said above,

--1

--2

3-6-4 and

--5

 

--6

--2

4-1 3

--5

 

Look like the same die. So, in reality, there are two answers to this problem- one in which the above to are the same, and one in which the above is a discrepancy ( for example, each side of the die is a different color and youre asking how many differences there are in that possibility.

 

In these situations, i find it best to stay away from factorials and random functions you learned in middle school that seem to always work but actually dont in most situations. Just take each space- the number of possibilities for each, and multiply. For case number two, in which the above examples are counted as different, the answer is 48. How?

 

Ok, so Cube. O's will represent open spaces and X's will represent Closed ones (while the others represent the number of possibilities)

 

--O

--O

.OOO

--O

 

The topmost space has six possibilities, so we then count

 

--6

--O

.OXO

--O

The center is an X because once one of the 6 numbers is chosen to be at the top, then there must be a certain number in the middle there. Afterwards we get

--6

--4

.OXO

--X

4 possibilities for the second number for four numbers remained. The final looks like

--6

--4

.2XX

--X

 

The number of possibilities is 6x4x2, which is 48.

 

However, if the difference stated above doesnt matter, then there are actually only 6 possibilities. If you don't believe me then i can draw it out, but the way i did it cant really be done with ascii.

[Jagged]

There is one unspecified thing about this problem. As said above,

--1

--2

3-6-4 and

--5

 

--6

--2

4-1 3

--5

 

Look like the same die. So, in reality, there are two answers to this problem- one in which the above to are the same, and one in which the above is a discrepancy ( for example, each side of the die is a different color and youre asking how many differences there are in that possibility.

 

In these situations, i find it best to stay away from factorials and random functions you learned in middle school that seem to always work but actually dont in most situations. Just take each space- the number of possibilities for each, and multiply. For case number two, in which the above examples are counted as different, the answer is 48. How?

 

Ok, so Cube. O's will represent open spaces and X's will represent Closed ones (while the others represent the number of possibilities)

 

--O

--O

.OOO

--O

 

The topmost space has six possibilities, so we then count

 

--6

--O

.OXO

--O

The center is an X because once one of the 6 numbers is chosen to be at the top, then there must be a certain number in the middle there. Afterwards we get

--6

--4

.OXO

--X

4 possibilities for the second number for four numbers remained. The final looks like

--6

--4

.2XX

--X

 

The number of possibilities is 6x4x2, which is 48.

 

However, if the difference stated above doesnt matter, then there are actually only 6 possibilities. (or at least im pretty sure, although others said two). I think its three because, well, think about it. Just think about the numbers 1, 2, and 3. On this die that is mentioned, the numbers 1,2,3 must ALL be present/touching/3d visible at one and only one vertice(sp?) of the cube (like, all must be present at some corner here http://home.cc.umanitoba.ca/~gunderso/model_photos/platonic/cube.jpg where one two and three are all seen). This must be true because on a cube the only two numbers that dont touch are ones opposite each other, and neither 1 nor 2 nor 3 is opposite to 1 , 2, or 3. Thus the only difference is where the 1, 2, and 3 are relative to eachother. 3 numbers, 3 places, 3!. 6

[woelen]

If the dice has unique sides (e.g. each side has its own specific color), then the situation is:

 

First number can be put on any of the 6 sides. This fixes the opposide side. Third number can be put on any of the 4 remaining sides, the fourth number and side are fixed again. Finally, the fifth number can be put on any of the 2 remaining sides.

 

Total is 6*4*2 = 48

 

If sides are not uniquely determined and rotated versions of the dice are considered equal, then initially, it does not matter where the first number is placed. In that way, 2 sides already are fixed. For the remaining four sides, it does matter where number 3 is placed, and all 4 sides are different (I do not assume that reflected dices are equivalent, consider them as left and right hands, which also are different). Finally, 2 sides remain for the fifth number and the sixth is fixed.

 

Now the total will be 4*2 = 8.

[uncool]

I see only two ways...

First, fix the 1 and 6 (one fixes the other) to be on top and bottom (we can always rotate the cube this way). Then fix 3 in front (once again, can be rotated to fit), and 4 is fixed. Now, there are 2 ways to fill the last 2, so there are a total of 2 different ways.

=Uncool-

[woelen]

The OP should come back on this and specify exactly what he wants.

 

If rotation is allowed, and also the orientation, relative to the painting on one side, does not matter, then indeed, there are only 2 variations, as uncool pointed out. If orientation, relative to the painting does matter, but rotation does not matter, then there are 8 variations, if rotations also is not allowed then there are 48 variations.

 

With how things are painted I mean the following. Suppose a 6 is placed on one of the sides, then we have

 

X
X6X
X

 

With the X's I mean the sides, around the side with the 6. Now, one can choose the 4 on one of the sides, and that fixes the 3.

 

4
X6X
3

 

and

X
463
X

 

are not equivalent, if one is interested in how things are painted. If e.g. the 6 is pained as two rows of 3 dots, then in one situation, the 4 and 3 are along the lines of the dots, otherwise they are perpendicular to the lines of the dots. If numbers are painted, then a similar thing is true.