Maximum number of students such that they all have at least 3 out of 6 different answers. A teacher made a test for his mathematics class with 6 true or false questions. When he received the tests, he noticed that any pair of students had at least three diferent answers. Since all the students answered every question, what's the maximum number of students of the class?
Here is my take. $2^6=64$ diferent tests. For one of those tests, there are $^6C_3+^6C_4+^6C_5+^6C_6$=42 tests which have 3 or more diferent questions from it. 
$42/63 = 2/3$, and if the ratio stands for successive pairs of tests, ${3/2}^{10}<63<{3/2}^{11}$ then 10 students. This argument is clearly fallacious and unfounded, but I found no better. 

To see that $9$ is impossible:
Note that if you only had $4$ questions you could have no more than $2$ students whose pairwise answers differed by at least three questions.  Indeed, pick one of the students.  Relabeling if necessary we can assume that this student chose $TTTT$.  Any other student would need to have chosen at least three $F's$, but then no any third student who chose at least three $F's$ would have to match the second in at least two places.
Now, take your proposed group of $9$ students.  Since there are exactly $4$ ways to respond to the first two questions, there must be at least $3$ students who match on the first two.  By the preceding comment, however, this is not possible.
