## Task 1 - 320731

a) Four red balls, two yellow balls, and one blue ball are to be distributed into two boxes $A$ and $B$ such that box $A$ contains three balls and box $B$ contains four balls.

How many such distributions are there in total?

b) Now, balls of the same color are distinguished by an additional numbering. The distributions then differ not only in how many balls of each color are in boxes $A$ and $B$, but also in which numbers they carry.

How many such distributions are there in total?
Every possible distribution is already uniquely determined by the numbers of balls in $A$. a) For these numbers, there are exactly the possibilities of the following table:

| Distribution No. | 1 | 2 | 3 | 4 | 5 | 6 |
| :--- | :--- | :--- | :--- | :--- | :--- | :--- |
| red | 0 | 1 | 1 | 2 | 2 | 3 |
| yellow | 2 | 1 | 2 | 0 | 1 | 0 |
| blue | 1 | 1 | 0 | 1 | 0 | 0 |

This results in a total of 6 distributions.

b) Each distribution to be determined can be obtained by identifying which numbers the balls can carry for each distribution from a):

Distribution No. 1 leads to exactly 1 distribution, as the only blue ball and all yellow balls must be in $A$.

For Distribution No. 2, any of the four red balls and any of the two yellow balls can be in $A$. This results in exactly $4 \cdot 2 = 8$ distributions.

For Distribution No. 3, all yellow balls must be in $A$, and any of the four red balls can be in $A$, resulting in exactly 4 distributions.

For Distribution No. 4, the only blue ball is in $A$, and for the numbers of the red balls, there are exactly the possibilities $(1; 2), (1; 3), (1; 4), (2; 3), (2; 4), (3; 4)$, leading to 6 distributions.

For Distribution No. 5: In $A$, any of these six combinations of red balls and any of the two yellow balls can be, resulting in exactly $6 \cdot 2 = 12$ distributions.

For Distribution No. 6: For the balls in $A$, exactly one of the four red balls must be omitted, resulting in exactly 4 distributions.

This results in a total of $1 + 8 + 4 + 6 + 12 + 4 = 35$ distributions.