Professor Doublebrain was severely ill last week, and he had to spend the six days from Monday till Saturday in the hospital. Luckily he has fully recovered by now. He told us that on each of these six days of illness he was visited by some of his 20 closest friends; and of course many of these friends visited him more than once. Professor Erasmus thought about this for some time, and then told me:

"No matter how and when and how often these 20 friends were visiting Professor Doublebrain on these six days of illness, one will be able to pick two days $D_1$ and $D_2$ and pick five friends $F_1,F_2,F_3,F_4,F_5$, such that the following holds: either all five friends visited Doublebrain on both days $D_1$ and $D_2$, or all five friends visited Doublebrain neither on day $D_1$ nor on day $D_2$."

Is this statement correct, or has professor Erasmus once again made a mathematical blunder?


1 Answer 1


If worded like this the solution is the one i tried to give on the first part.

Six days of illness (Part 1)

Professor Erasmus is wrong

Proof by Example:

|  D   Friends--------------->
|  a
|  y
\/ s

1111 1111 1100 0000 0000
1111 0000 0011 1111 0000
1000 1110 0011 1000 1110
0100 1001 1010 0110 1101
0010 0101 0101 0101 1011
0001 0010 1100 1011 0111

I did this by stating all 20 permutations of exactly 3 visits on 6 days.
You will find no 2 days with either more than 4 Friends visiting or more than 4 friends not visiting.


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