Mathematics-Online problems: Solution to the problem of the (previous) week

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	Mathematics-Online problems:
	Solution to the problem of the (previous) week

Problem:

#./interaufg22_en.tex#On a die the numbers 1 to 6 are depicted by the common symbols. Two dice are considered equal, if they can be rotated in space in such a way that corresponding faces show the same symbol in the same arrangement. Accordingly, the dice shown below are thus all different.

$\includegraphics{wuerfel}$

How many different dice are possible, if the spots on opposite faces add up to seven, as is commonly demanded?
How many different dice are possible, if there is no such restriction?
How many different dice are possible, if it is demanded that there is at least one pair of opposite faces whose spots add up to nine?

Answer:

Number of different dice, if

$\bullet$ opposite faces add up to 7 spots

$\bullet$ opposite faces add up to any number of spots

$\bullet$ at least one pair of opposite faces adds up to 9 spots

Solution:

When considering the various possibilities of depicting the numbers 2, 3, and 6, it is evident that the number of differing dice increases eightfold.

Spots on opposite faces add up to 7
In any case, 1 and 2 are on neighbouring faces. Consequently, each die can be turned in such a way that 1 spot is on top and 2 spots are on the front face. Then, 6 spots are on the bottom and 5 spots on the back.
The positions of 3 and 4 spots are interchangeable.
Thus, without regarding the arrangement, there are possibilities; that is a total of $2 \cdot 8 = 16$ possibilities.
Spots on opposite faces add up to any number
We obtain a total number of possibilities when considering the following:
- Without loss of generality, there is 1 spot on the top face.
- Then, possibilities remain for the number of spots on the bottom face.
- The remaining four numbers of spots can be distributed in ways. However, arrangements can be transferred into each other rotating the die around the vertical axis.
Spots on at least one pair of opposite faces add up to 9
We have to take into consideration that
- the opposite face of 6 must have 3 or the opposite face of 5 must have 4 spots;
- in both cases, there are again possibilities for the remaining four numbers of spots (see above);
- if 3 spots are opposite 6 and 4 spots opposite 5, there are still possibilities for 1 and 2 spots.
Altogether, there are different dice.

[problem of the week]

Number of different dice, if
$\bullet$	opposite faces add up to 7 spots
$\bullet$	opposite faces add up to any number of spots
$\bullet$	at least one pair of opposite faces adds up to 9 spots