In how many ways can 1 apple, 2 peaches, 3 cucumbers, and 2 plums be distributed among Annus, Bözsi, Cili, and Dóra so that each girl gets 2 whole fruits? (Fruits of the same type cannot be distinguished from each other.)
Solution. 1. We will provide the answer in two steps, and as preparation, we will list the ways in which 8 fruits can be arranged on 4 identical plates, with 2 fruits on each. We will call our arrangement principle that we fill the plates one after the other, and if during this process we see multiple ways to continue, we proceed by first using the type of fruit of which there is currently the least; if there are multiple types of fruits with the same least quantity, we proceed in alphabetical order of their initials. We will denote the fruits by their initial letters.

The possibilities for filling the first plate:
a) $a, b$,
B) $a, s z$,
$\gamma$ ) $a, k$

And in the $\alpha$) case, the following 2 situations can be created on the first two plates:

$\left.\left.\alpha_{1}\right) \quad a, b, \quad b, s z ; \quad \alpha_{2}\right) \quad a, b, \quad b, k$.

The former situation clearly leads to the arrangement in the following table (I), while the latter can be completed in two ways, because there are 2 each of $k$ and $s z$ left: by placing only one type of fruit on each of the last two plates, or mixed: (II) and (III). (For simplicity, we will no longer insert commas between the members of a pair of letters.)

The continuations of the $\beta$) start can be obtained from (I)-(III) by swapping the $b, s z$ letters, since their quantities in the stock are equal, these are the (IV)-(VI) arrangements, which are clearly different from the former, and since the first three are different, the latter three are also different.

Applying our principles to $\gamma$), we get the (VII)-(XI) plate fillings.
(I) $a b, b s z, s z k, k k$;
(IV) $a s z, s z b, b k, k k$
(II) $a b, b k, k k, s z s z$
(V) $a s z, s z k, k k, b b$
(III) $a b, b k, k s z, k s z$
(VII) $a k, b b, k k, s z s z ;$
(VIII) $a k, b b, k s z, k s z$
(IX) $a k, b k, b k, s z s z$
(X) $a k, b k, b s z, k s z$
(XI) $a k, b s z, b s z, k k$.

2. In the (I), (II), (IV), (V), (VII), and (X) arrangements, there are no two plates filled identically, while in the other five, the exchange of 2-2 identical plates is not noticeable. In the former 6 arrangements, we can choose 4 ways to decide which girl gets the first plate (serving), 3 ways from the remaining girls for the second plate, 2 ways for the third, and the fourth plate is uniquely determined by the one who has not yet received. Thus, we get $4 \cdot 3 \cdot 2=24$ distribution methods from each of the 6 cases, totaling 144.

Considering the 24 distribution methods for each of the other 5 arrangements, they yield identical results in pairs, so only 12 are different, for example, if in (III) we give the third plate to Cilin and the fourth to Dora, the distribution does not change if these two (indistinguishable) servings are swapped between them.

Accordingly, the 5 arrangements containing 2-2 identical servings give $5 \cdot 12=60$ distribution methods, and the answer to the question is: $144+60=204$ different distributions are possible.

Remarks. 1. Another starting point for compiling the (I)-(XI) arrangements: the 3 apples go into either 2 or 3 different plates. In the former case, the quantities to be further divided are $1,2,1,2$, and with our principle, $a k, b b, s z s z$ (VII), $a k, b s z, b s z$ (XI); $a b, b k, s z s z$ (II); $a b, b s z, k s z$ (I); and (with $b, s z$ swapped) $a s z, k s z, b b$ (V); $a s z, k b, s z b$ (IV); result. In the latter case, we can choose 5 ways for the still empty plate from the quantities $1,2,0,2$: $a b$ (III); $a s z$ (VI); $b b$ (VIII); $b s z(\mathrm{X}) ; s z s z$ (IX), and the remaining 3 fruits can be placed one by one on the still equal (not distinguished) plates through their 1-1 apple.

2. We can also organize according to whether we prepare $3,2,1$ or 0 servings that contain only one type of fruit; the groups receive $1,2,5$ and 3 arrangements.