A class has 42 female students whose three favorites are: an actor ($S z$), a fencer $(V)$, and a skater $(K)$. The girls collect signed photos of their favorites. Six girls have photos signed by all three. The class collectively has 21 $S z$ photos, 20 $V$ photos, and 18 $K$ photos. When asked who has both an $S z$ photo and a $V$ photo, 7 girls responded, similarly, 10 for $S z$ and $K$, and 11 for $V$ and $K$. According to this, is there a girl who does not have a signed photo from any of the favorites, and if so, how many are there?
Note. In the solution, we assume that a girl does not have two photos of the same favorite. Several correctly pointed out that without this, the problem cannot be solved unambiguously.

I. Solution. Draw two intersecting circles, and another one that covers part of their common area but not all of it. Thus, part of the common area of any two circles is inside the third circle, while another part is outside. Let's name the circles $S z$, $V$, and $ $K$ respectively.

![](https://cdn.mathpix.com/cropped/2024_05_02_e3b568516b34754daad6g-1.jpg?height=707&width=553&top_left_y=411&top_left_x=775)

Now, place each girl in such a way that she stands in every circle of her favorite for which she has a photo. If she does not have a photo of a certain favorite, she stands outside that circle. If there is a girl who does not have any photos, she stands outside all three circles, and we need to determine the number of such girls. To do this, we calculate the number of girls in each defined region.

In the common area of the $S z$ and $V$ circles, 7 girls stand, of which 6 are also in the part that belongs to $K$, so 1 stands in the part outside $K$. Similarly, in the part of the common area of the $S z$ and $K$ circles that does not belong to $V$, $10-6=4$ girls stand, and $11-6=5$ girls stand in the part of the common area of the $V$ and $K$ circles that does not belong to $S z$.

The remaining regions belong to only one circle. In the $S z$ circle, 21 girls stand, of which $1+6+4=11$ are in parts that also belong to other circles, so 10 stand only in $S z$. Similarly, 8 girls stand only in $V$, and 3 girls stand only in $K$.

In the 7 regions, 37 girls stand in total, leaving 5 outside, meaning these 5 do not have any signed photos.

$$
\text { Éva Krassai (Székesfehérvár, Teleki B. lg. II. o. t.) }
$$

Note. There are 8 possibilities for which photos a girl has and which she does not. Specifically, whether $S z$ is present or not (2 possibilities), each of these splits into two cases depending on whether $V$ is present or not, and each of these 4 possibilities further splits depending on whether $K$ is present or not. These 8 possibilities can be represented by the 8 vertices of a cube. The number of girls in each case can then be determined, for example, by placing the symbols of those who have $S z$ on the first face of the cube, and those who do not on the opposite face. From these, move the symbols of those who have $V$ to the right edge, and those who do not to the left edge. Finally, on each edge, move the symbols of those who have $K$ to the upper vertex, and the rest to the lower vertex. Following this procedure, we essentially arrive at the previous solution.

This method can also make comprehensible cases where a girl can have multiple photos of the same favorite. Accordingly, for example, starting from the back face, parallel cross-sections should be inserted for the symbols of those who have 1, 2, 3, etc., $S z$ photos, and similarly, parallel cross-sections would be needed for the other faces (and, of course, more data than in the problem).

II. Solution. The problem can also be solved without calculating the number of girls for each possible collection of photos, as we did above.

If every girl had at most one photo of a favorite, we could find the number of girls without photos by subtracting the total number of photos from the class size. If we calculate this way in our case, we get a negative result:

$$
42-(21+20+18)
$$

This is understandable because we subtracted the number of those who have more than one photo multiple times. For example, we subtracted twice the number of those who have both $S z$ and $V$ photos, i.e., in both 21 and 20. We can correct all such double subtractions by adding the total number of those who appeared for the pairs of photos $S z-V$, $S z-K$, and $V-K$ to the above difference:

$$
42-(21+20+18)+(7+10+11)
$$

Furthermore, we subtracted three times the number of girls who have all three photos. However, we already added their number three times because they obviously appeared for all three pairs of photos. We need to subtract these 6 girls again, so the number of girls who do not have any photos is

$$
42-(21+20+18)+(7+10+11)-6=5
$$

Miklós Mátrai (Hódmezővásárhely, Bethlen G. g. II. o. t.)