Combinatorics Through Guided Discovery, 2004a

174 D. Hints to Selected Problems
67. What usually makes it hard for students to start this problem is the fact that
we just defined what R(4, 4) is, and not what it means for a number not to be
R(4, 4). So to get started, try to write down what it means to say R(4, 4) is
not 8. You will see that there are two things that can keep R(4, 4) from being
8. You need to figure out which one happens and explain why. One such
explanation could involve the graph K 8 .
68. Review Problem 65 and your solution of it.
69. Let a i be the number of acquaintances of person i. Can you explain why the
sum of the numbers a i is even?
70. Often when there is a counter-example, there is one with a good deal of
symmetry. (Caution: there is a difference between often and always!) One
way to help yourself get a symmetric example, if there is one, is to put 8
vertices into a circle. Then, perhaps, you might draw green edges in some
sort of regular fashion until it is impossible to draw another green edge
between any two of the vertices without creating a green triangle.
71. InProblem 68 you showed that R(4, 3) ≤ 10. InProblem 70 you showed that
R(4, 3) > 8. Thus R(4, 3) is either 9 or 10. Deciding which is the case is just
plain hard. But there is a relevant problem we have done that we haven’t
used yet.
72. We wish to prove that ( n i ) = n!
i!(n−i)!
. Mathematical induction allows us to
assume that ( n−1 )
j
= (n−1)!
j!(n−1−j)!
for every jbetween 0 and n − 1. How does this
put us into a position to use the Pascal relation? What special cases will be
left over?
73. What sort of relationship do you know between values of the form ( n i ) and
values of the form ( n−1
j
)?
75. We did something rather similar in our example of the inductive proof that a
set with n elements has 2 n subsets. The work you did in a previous problem
may be similar to part of what you need to do here.
76.a. This may look difficult because one can’t decide in advance on whether to
try to induct on m, onn, or on their sum. In some sense, it doesn’t matter
which you choose to induct on, though inducting on the sum would look
more complicated. For most people inducting on n fits their way of working
with exponents best.
76.b. Here it matters whether you choose to induct on m or n. However, it matters
only in the sense that you need to use more tools in one case. In one case, you
are likely to need the rule (cd) n = c n d n (, which we haven’t proved. (However,
you might be able to prove that by induction!) In either case, you may find
part (a) handy.