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whats so lol
R u sure it cld be proved
of course it is
oh i see didnt read the question correctly
it's classical problem and i have proved it somehow but i don't remember anymore
do the 6 people even have to be friends? after all, a group of 6 strangers is still a group of 6
no, but if they don't have any friends there will be 6 people who have no friends
if 2 people in the group are friends; then at least 2 people have 1 friend .... something like that
if 3 people are friends, then a > bc, b > ac, c>ba
thats at least 2 with the same amount of friends
or even a>c , c>b, c>ab
well i translated it from lithuanian but it seems it has same meaning
can 2 people be not friends? if one of them is a friend?
oh wait, I see. Let f(n) be the number of friends in that group that the nth person has. For each n, \[ 0 \leq f(n) \leq 5 \] Therefore ...
a>c, b>c , c>ab is what i meant lol
yes there can be no friends at all or 1 person can have no friends
I think James has it :)
even if 4 are friends, that still leaves 2 with 0 friends each which is the same amount
there are 6 values for f(1), f(2), f(3), f(4), f(5), f(6). Now it can't be that these six number take on all of 0,1,2,3,4,5. Because if f(n) = 5 for one n, then f(j) > 0 for all other \( j \neq n \). Hence in fact, there are only 5 possible values for f(n). But as there are six f(n), at least two of the f(n) must be equal.
ill concede to that answer as well ;)
my by case proof could get lengthy
gotta hang it on the left to be a valid q lol
and I gotta get my ode hw written up so ciao yall :)