5. In how many ways can Iva, Ana, and Tea divide 6 different awards so that each of them gets at least one award?

The use of a pocket calculator or any manuals is not allowed.

## Ministry of Science and Education of the Republic of Croatia  Agency for Education and Upbringing  Croatian Mathematical Society

## COUNTY COMPETITION IN MATHEMATICS  March 24, 2022.
First solution.

From the conditions of the problem, it is clear that all prizes must be distributed, but not all girls have to receive the same number of prizes.

We distinguish three cases: 1-1-4, 2-2-2, 1-2-3,

1 POINT where the numbers indicate how many prizes a person receives.

In each case, we need to determine how many ways we can choose which girl will receive which number of prizes, and which specific prizes they will receive.

$1^{\circ}$ Case $1-1-4$

We can determine which girl receives four prizes in three ways.

1 POINT

Suppose Iva and Ana each receive one prize, and Tea receives four prizes. Iva's prize can be chosen in 6 ways, Ana's in 5 ways, and for Tea, the remaining 4 prizes can be chosen in only one way. This gives us $6 \cdot 5 \cdot 1=30$ ways.

Similarly, we conclude in the remaining two cases, so the total number of ways in this case is $30 \cdot 3=90$.

Iva's two prizes out of the total 6 prizes can be chosen in $(6 \cdot 5): 2=15$ ways, Ana's two out of the remaining 4 prizes can be chosen in $(4 \cdot 3): 2=6$ ways, and finally, for Tea, the remaining two prizes can be chosen in one way.

Thus, the number of ways to distribute the prizes in this case is $15 \cdot 6 \cdot 1=90$. out of those five prizes can be chosen in $(5 \cdot 4): 2=10$ ways, and for Tea, the remaining three prizes can be chosen in one way. Thus, the number of ways to distribute the prizes where Iva receives one, Ana receives two, and Tea receives three prizes is $6 \cdot 10 \cdot 1=60$.

Note: If a student obtained the total number of prize distributions as the sum $90+270+180=540$, they can receive a maximum of 7 POINTS.