Earlier I provided an "answer" as a hint (in the form of a question). It was deleted as being more appropriate as a comment ("request for clarification") it was not a request for clarification. It was a hint in the form of a question.
"Hint: How many total A's were there out of how many total grades?"
The point is that there were 
total A's out of 
total grades. Hence there were 
grades that were not A's. If these grades were maximally spread around among the 
students, there would have been 
students who got a non A. (There may have been less such students, as some students could have gotten more than one non A.) But in any case that would mean that there are least 
students who got all A's.
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How do I calculate the intersection between three sets?
To calculate the intersection between three sets, A, B, and C, you can use the inclusion-exclusion principle. We assume you will know all the quantities mentioned.
- Find the cardinality of the union of all three sets (
|A ∪ B ∪ C|). - Subtract the cardinality of every single set(
|A|,|B|, and|C|). - Add the cardinality of the intersection of each pair of set (
|A ∩ B|,|A ∩ C|, and|B ∩ C|).
The result will be the cardinality of the intersection |A ∩ B ∩ C|.
What is the symmetric difference of two sets?
The symmetric difference of two sets, A and B, is the set that contains all the elements belonging exclusively to either A or B. The corresponding logical operation is the exclusive or: when both sets exist in a given portion of the diagram, we disregard it. With higher numbers of sets in your diagram, use sum modulo 2 of the overlaps of the sets. If the overlaps are even, exclude the subset. If the overlaps are odd, include the subset.
What is the union if |A| = 10, |B| = 12, and |A ∩ B| = 4?
The union of A and B, with |A| = 10 and |B| = 12 is |A ∪ B| = 18. To find this result, we use the inclusion-exclusion principle:
- Calculate the sum of the cardinalities
|A|and|B|:|A| + |B| = 10 + 12 = 22. - Subtract the cardinality of their intersection:
|A| + |B| - |A ∩ B| = 22 - 4 = 18.
That's it. We subtracted the intersection because, by summing |A| and |B|, we inadvertently counted twice the elements shared by the two sets.
Earlier I provided an "answer" as a hint (in the form of a question). It was deleted as being more appropriate as a comment ("request for clarification") it was not a request for clarification. It was a hint in the form of a question.
"Hint: How many total A's were there out of how many total grades?"
The point is that there were total A's out of total grades. Hence there were grades that were not A's. If these grades were maximally spread around among the students, there would have been students who got a non A. (There may have been less such students, as some students could have gotten more than one non A.) But in any case that would mean that there are least students who got all A's.
Let 
be the number of students that got all As.
We use the inclusion-exclusion formula,
to get
To find a bound, we need to bound the intersections: using 
we can rearrange to obtain 
Plugging that (and analogous expressions) back into the equality for 
gives
Thus 
, because 
.