CS 2800 – Fall 2008
Homework #6 Solution
Section 5.1
Problem 32 [4pts]
For each element in the domain, we have
n
elements to choose from in the codomain. This
choice needs to be repeated 10 times, for each element in the domain.
(a) 2
10
(b) 3
10
(c) 4
10
(d) 5
10
Problem 38 [4pts]
Number of subsets with more than one element = Total number of subsets  Number of
subsets with one element  Number of subsets with no elements = 2
100

100

1 = 2
100

101.
Section 5.2
Problem 16 [5pts]
We first group the numbers in the given set into four pairs that add up to 16 each, namely
{
1
,
15
}
,
{
3
,
13
}
,
{
5
,
11
}
,
{
7
,
9
}
. Now we see that if we were to pick 5 numbers from the original
set, then by the pigeonhole principle, two of them must belong to the same pair, and hence
we are guaranteed at least one pair of numbers that add up to 16.
Problem 26 [5pts]
We begin by noting that even though the result from exercise 25 applies to groups of size
exactly
10, it also applies to groups of size greater than 10 because we can simply remove
individuals from the group until the group size reaches 10 and then use the result. Adding
back the removed individuals does not destroy the preexisting cliques of friends/enemies.
Now, pick an arbitrary person
x
in the group of 20 people. By definition,
x
is related to
all the other 19 people in the group, by either a ‘friend’ link or an ‘enemy’ link. Since the
links come in only two flavours, by the pigeonhole principle, we can conclude that
x
has at
least
19
2
= 10 links of the same type i.e.
x
either has at least 10 friends or 10 enemies.
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 '07
 SELMAN
 Combinatorics, RHS

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