[home] [lexicon] [problems] [tests] [courses] [auxiliaries] [notes] [staff] | ||

Mathematics-Online problems: | ||

## Solution to the problem of the (previous) week |

** Problem:**

- 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 | ||

opposite faces add up to 7 spots | ||

opposite faces add up to any number of spots | ||

at least one pair of opposite faces adds up to 9 spots |

** Solution:**

**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 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.

[problem of the week]