Looks like some context is missing.
This looks like an explanation on how to count elements in the union of 3 sets A, B, and C.
Btw, the notation |A| means the number of elements in A (also called the cardinality of A).
The formula for that is:
$$|A\cup B \cup C| = |A|+|B|+|C|- \Big( |A \cap B|+|A \cap C| +|B \cap C| \Big)+|A \cap B \cap C|$$
In words: the total number of elements in the union is the sum of the elements in each set minus the elements in the mutual intersections plus the elements in the 3-way intersection.
I believe the rightmost drawing represents the actual situation.
That is, we have 3 sets A, B, and C that each have 4 elements, such that each of the different types of intersections contain 1 element.
The total number of elements in the union is 7.
The leftmost drawing shows what you would count if you calculate |A|+|B|+|C|.
In that case each element in the intersections is counted twice, hence the 2 in the overlaps.
Except for the part where all 3 sets intersect where each element is counted thrice, hence the 3 in the middle.
The middle drawing represents what you get with only the part of the formula that is given below it.
The rightmost drawing is putting everything together, effectively showing the original situation.
It seems it's the case that circles are filled up with numbers (look at numbers as objects like tomatoes!) and the image is asking you find solve the desired equations. Just you should decide at first the sets A, B and C in the image!! unless you would not be able to solve it. For example, for the leftmost picture, let the up-left circle be A, the up-right one be the set B, and the down one be C, the the answer of the equation |A|+|B|+|C| is equal to 4+4+4=12.
And the other possibility is what bergausstein mentioned! (And You again need to mark the sets first!)
The numbers indicate how many times that region has been counted using the equation below it. In the first one, you have three regions which are double counted and one region that is triple counted so you then subtract out $\Big( |A \cap B|+|A \cap C| +|B \cap C| \Big)$, which gets you close. We see that after doing this all regions are counted once, as desired, except for the middle region so we add back in $|A \cap B \cap C|$ and now we correctly are counting each region just one time.