Professor Carlão decided to create a math question worth a total of 10 points and consisting of three items: $a, b$, and $c$. After formulating the items, he became unsure about the best way to distribute the 10 points among the items so that each one is worth a positive integer number of points.

a) Joana, a teacher friend of Carlão, suggested that item $c$ should be worth the same number of points as the sum of items $a$ and $b$ because, according to her, item $c$ is more difficult. If Carlão follows Joana's suggestion, in how many different ways can he distribute the points?

b) Disregarding Joana's suggestion, that is, considering that Carlão will distribute the points in any way, in how many different ways can he distribute the 10 points of the question among the three items?
Solution

a) If Carlão follows Joana's suggestion, item $c$ will be worth 5 points and items $a$ and $b$ must sum to another 5 points. We will then have four possible divisions of items $(a, b, c):(1,4,5),(2,3,5)$, $(3,2,5)$, and $(4,1,5)$.
b) Once the scores for items $a$ and $b$ are defined, item $c$ will be worth $10-a-b$ points, and therefore, we only need to count the number of ways to choose $a$ and $b$. If items $a$ and $b$ together are worth $n$ points, then we will have $n-1$ possibilities for pairs of positive integers $(a, b):$

$$
(1, n-1),(2, n-2),(3, n-3), \ldots,(n-1,1)
$$

Since the sum $a+b$ must be at most 9, when $c$ is at its minimum and worth 1 point, and at least $1+1=2$ points, when $a$ and $b$ are at their minimums, the total number of ways to distribute these points is

$$
\begin{array}{r}
(9-1)+(8-1)+(7-1)+(6-1)+(5-1)+(4-1)+(3-1)+(2-1)= \\
8+7+6+5+4+3+2+1=
\end{array}
$$

36.

Thus, Carlão can distribute the 10 points in 36 different ways.