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Computer Science 20
In-Class Problems 25
Monday, April 4, 2016
Author: Jack Dent
Executive Summary
1. Fundamental rules for counting
• The sum rule: If sets A and B are disjoint, A has a elements, and B has b elements,
their union has a + b elements. This permits a “divide and conquer” strategy.
• The difference rule: Suppose set B has b elements and its subset A has a elements. Then
the set difference B − A has b − a elements. This rule is useful when it is easy to count
sets B and its subset A, but there is no simple way to count B − A directly.
• The product rule: Suppose that set A has m elements and set B has n elements. The
Cartesian product A × B is the set of ordered pairs (a, b), where a is an element of A
and b is an element of B. The set A × B has mn elements.
• The power rule: If A has n elements and you select a sequence of r elements with
replacement from A, there are nr possible sequences.
• The division rule: When counting, it is fine to count things more than once, provided
that everything gets overcounted the same number of times and that you know that
number and divide by it.
PROBLEM 1
Twelve athletes are competing in a race.
(A) How many possibilities are there for the podium positions (first, second, and third place)?
(B) Suppose now that two people tie in first place, three people tie in second place, and 2 people
tie in third. How many different possibilities for the podium are there now?
Solution.
(A) Ordering the athletes and dividing out the irrelevant positions, we have 12!
9! .
(B) Dividing by the irrelevant positions, and dividing by the number in each podium position since
12!
we don’t care about ordering, we have 2!·3!·2!·5!
.
PROBLEM 2
Compute the number of anagrams for each of the following words. Assume that anagrams need
not be proper words.
(A) “lewis”
(B) “engineer”
(C) “mathematician”
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Solution.
(A) There are no repeated characters, so there are 5! anagrams.
8!
(B) There are two ‘n’s and three ‘e’s, so there are 2!·3!
anagrams.
(C) There are two ‘m’s, three ‘a’s, two ‘t‘s and two ‘i’s, so there are
13!
2!·3!·2!·2!
anagrams.
PROBLEM 3
(A) A dodecahedron has 12 faces, each a regular pentagon. How many edges does it have?
(B) Three faces of the dodecahedron meet at each vertex. How many vertices does the dodecahedron
have?
Solution.
(A) Each face has 5 edges, and each edge is shared by two faces, so there are 12×5
2 = 30 edges.
(B) Each face has 5 vertices, and each vertex is shared by three faces, so there are 12×5
= 20
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vertices.
PROBLEM 4
A license plate consists of either
• 3 letters followed by 3 digits (standard plate)
• 5 letters (vanity plate)
• 2 characters - letters or numbers (big shot plate)
Compute the number of different possible license plates.
Solution.
263 × 103 + 265 + 362
PROBLEM 5
How many of the billion numbers in the range from 1 to 109 (inclusive) contain the digit 1?
Solution.
By the product rule, 99 of the numbers between 0 at 109 − 1 do not contain a 1 and neither does
109 , so there are 109 − (99 − 1) numbers that do contain the digit 1.
PROBLEM 6
(BONUS) Vandermonde’s Identity states that:
n X
r
s
r+s
·
=
k
n−k
n
k=0
2
Give an argument for why both sides of the equation are equivalent.
Solution.
Suppose we have two groups of people,
R and S with |R| = r and |S| = s. We can select a group
ways, which is the right side of the equation. Now suppose
of n people from the set R ∪ S in r+s
n
we select k people from set R, and the remaining n − k from
set S. Multiplying the number of
s
possible combinations in both sets, we can do this in kr · n−k
ways. Sumimg over all k between 0
and n inclusive, which is the left side of the equation, we obtain the total number of ways to select
n people across both sets, and so the identity holds.
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