The figure below indicates a target fixed on a wall that cannot be rotated. It is divided into 10 parts, consisting of a central circle, a smaller ring, and a larger (outer) ring. We need to distribute the numbers from 1 to 10, one in each part, which will correspond to the scores obtained by hitting each part.

![](https://cdn.mathpix.com/cropped/2024_05_01_a50b5476db779c3986d5g-07.jpg?height=497&width=510&top_left_y=1940&top_left_x=845)
a) In how many ways can we distribute the numbers in the parts of the target?

b) In how many ways can we distribute the numbers such that numbers closer to the center cannot be smaller than numbers more distant from the center?

c) In how many ways can we distribute the numbers so that the sum of the numbers in the outer ring is equal to the sum of the numbers in the smaller ring?
Solution

a) Since there are 10 parts and 10 numbers, the total number of possibilities is

$$
10!=10 \cdot 9 \cdot 8 \cdot \ldots \cdot 1=3,628,800
$$

b) In the central circle, the number must be 10 (only one possibility); in the smaller ring, the numbers must be 7, 8, and 9 for 3 parts $(3 \cdot 2 \cdot 1=6$ possibilities); in the outer ring, there are the remaining 6 numbers for 6 parts $(6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=720$ possibilities). Therefore, the total number of ways for this distribution is $1 \cdot 6 \cdot 720=4,320$.

c) The sum of all the numbers is $1+2+3+\ldots+10=55$ (odd). Since the sum of the smaller and outer rings must be the same, the two together must result in an even number, and consequently, the central circle must contain an odd number. Let's analyze each case:
I) 1 in the central circle (sum of each ring equal to 27): in the smaller ring, we must have 10, 9, and $8 (3 \cdot 2 \cdot 1 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=4,320$ possibilities $)$;

II) 3 in the central circle (sum of each ring equal to 26): in the smaller ring, we must have 10, 9, and $7 (3 \cdot 2 \cdot 1 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1=4,320$ possibilities);

III) 5 in the central circle (sum of each ring equal to 25): in the smaller ring, we must have 10, 9, and 6 or 10, 8, and 7 ( $2 \cdot 4,320=8,640$ possibilities);

IV) 7 in the central circle (sum of each ring equal to 24): in the smaller ring, we must have 10, 9, and 5 or 10, 8, and $6 (2 \cdot 4,320=8,640$ possibilities $)$;
V) 9 in the central circle (sum of each ring equal to 23): in the smaller ring, we must have 10, 8, and 5 or 10, 7, and $6 (2 \cdot 4,320=8,640$ possibilities $)$.

Therefore, there are $8 \cdot 4,320=34,560$ possibilities in total.