3. In how many different ways can four friends Ankica, Gorana, Ivana, and Melita divide 11 identical fries among themselves, so that each gets at least two? The fries cannot be cut into smaller pieces.
First solution.

Since each friend gets at least two croquettes, the number of croquettes remaining after this distribution is $11-4 \cdot 2=3$.

These three remaining croquettes can be distributed so that one friend gets all three, two friends share them, or three friends share them.

1) Three friends get 2 croquettes each, and one of them gets $2+3=5$ croquettes.

This can be done in four different ways:

| Ankica | Gorana | Ivana | Melita |
| :--- | :--- | :--- | :--- |
| 5 | 2 | 2 | 2 |
| 2 | 5 | 2 | 2 |
| 2 | 2 | 5 | 2 |
| 2 | 2 | 2 | 5 |

2) Two friends get 2 croquettes each, one gets $2+1=3$ croquettes, and one gets $2+2=4$ croquettes.

This can be done in twelve different ways:

| Ankica | Gorana | Ivana | Melita |
| :--- | :--- | :--- | :--- |
| 2 | 2 | 3 | 4 |
| 2 | 2 | 4 | 3 |
| 2 | 3 | 2 | 4 |
| 2 | 3 | 4 | 2 |
| 2 | 4 | 2 | 3 |
| 2 | 4 | 3 | 2 |
| 3 | 2 | 2 | 4 |
| 3 | 2 | 4 | 2 |
| 3 | 4 | 2 | 2 |
| 4 | 2 | 2 | 3 |
| 4 | 2 | 2 | 2 |
| 4 | 3 |  |  |

Such a distribution can be done in four different ways.

| Ankica | Gorana | Ivana | Melita |
| :--- | :--- | :--- | :--- |
| 3 | 3 | 3 | 2 |
| 3 | 3 | 2 | 3 |
| 3 | 2 | 3 | 3 |
| 2 | 3 | 3 | 3 |

In total, there are $4+12+4=20$ different ways that satisfy the conditions of the problem.

$1 \text{ POINT}$

Note: For every four correctly written distributions, 1 point is awarded. This means that if a student lists all possibilities and states that there are 20, but does not provide an explanation for why there are no more, they can receive a maximum of 6 POINTS.

TOTAL 10 POINTS

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