CS173 Discrete Mathematical Structures
Fall 2006
Homework #9 Solution
due Tuesday, November 7, 2006, 8:00 a.m.
(46 points + 6 bonus points)
1. The Pigeonhole Principle (10 points)
a. 5 points.
2 points for correctly grouping addresses into pairs of consecutive numbers.
2 points for applying the PHP.
1 point for including the fact that each house has a distinct address
Group 50 addresses into 25 pairs of consecutive numbers: {1000, 1001}, {1002,
1003}, …, {1048, 1049}. Now we assign 26 houses to 25 groups. According to the PHP
and the fact that each house has a distinct address, there must be two houses assigned to
the same group of addresses. Therefore, these two houses have consecutive addresses.
b. 5 points.
1 point for using only starting and ending letters of strings.
2 points for finding all combinations of the starting and ending letters.
2 points for using the PHP.
Nonempty strings over {a, b, c, d} can be any size greater than 0, can use any
combination of letters, and can have any order. We are only concerned with the starting
and ending letters of strings. Find all 16 possible combinations of the starting and ending
letters:
a*a a*b a*c a*d
b*a b*b b*c b*d
c*a c*b c*c c*d
d*a d*b d*c d*d
Now that we have 16 “pigeonholes” and 17 “pigeons”, we apply the PHP and conclude
that there must be two different strings that have similar starting and ending letters.
2. Combinatorial Proof (8 points)
4 points for each side. Deduct 6 points off for only rephrasing the hints.
First note that we can start the summation with k = 1, since the term with k = 0 is 0. The
left-hand side counts the number of ways to choose a subset as described in the hint by
breaking it down by the number of elements in the subset, note there are k ways to choose
each of the distinguished elements if the subset has size k. For the right-hand side, first note
that 2221
(1)2 (12)2 (1)2 2
nnnn
nn nn nn n
−−−−
+=−+ =−+. The first term counts the
number of ways to make this choice if the two distinguished elements are different (choose
them, then choose any subset of the remaining elements to be the rest of the subset). The
second term counts the number of ways to make this choice if the two distinguished elements
are the same (choose it, then choose any subset of the remaining elements to be the rest of the
subset). Note that this works even if n = 1.
