3. how many ways can you assign one of the numbers $1,2,3, \ldots, 10$ to each corner of a cube so that no number is used more than once and so that for each side face the sum of the numbers in the four adjacent corners is odd?
## Solution:

We call these sums of four numbers in the corners of a side surface area sums for short. Whether an area sum is odd or not depends only on how many of the numbers in the corners are odd. For each area, therefore, either 1 or 3 adjacent corner numbers must be odd.

Now fix the cube and look at a pair of opposite faces. If both faces have only one adjacent odd corner number, then a total of 6 corner numbers must be even, in contradiction to the fact that only five of the numbers $1,2, \ldots, 10$ are even. If both faces have three adjacent odd corner numbers, then a total of 6 corner numbers must be odd, again a contradiction. Testing the remaining four cases shows that the arrangement of the four odd corner numbers is as follows: One corner and the three adjacent corners have odd numbers, the rest have even numbers.

There are now 8 ways to choose the central odd corner. Subsequently, different numbers from the set $\{1,3,5,7,9\}$ can be assigned pairwise to the four odd corners in $5 \cdot 4 \cdot 3 \cdot 2$ ways. Assign different numbers from the set $\{2,4,6,8,10\}$ in pairs to the four even corners in $5 \cdot 4 \cdot 3 \cdot 2$ ways. The required number of assignments is therefore equal to

$$
8 \cdot(5 \cdot 4 \cdot 3 \cdot 2)^{2}=115200
$$