Lecture 2

The Counting Principle: Consider a process of $r$ stages. Suppose that:

There are $n_1$ possible results at the first stage.
For every possible result at the first stage, there are $n_2$ possible results at the second stage.
More generally, for any sequence of possible results at the first $i - 1$ stages, there are $n_1$ possible results at the $i^{th}$ stage.

Then, the total number of possible results of the r-stage process is $n_1n_2 \cdots n_r$

Big Idea: For problems without permutation or replacement, a series of decisions can be enumerated by multiplying the number of possible decisions at each stage.

Example: A pixel in a digital image is represented by 8 bits. How many encodings are possible for one pixel?

$2^8 = 256$

Mutually Exclusive Counting: Suppose that $X_1, \dots, X_t$ are sets and that the $i^{th}$ set $X_i$ has $n_i$ elements. If ${X_1 \dots, X_i}$ is a pairwise disjoint family, the number of possible elements that can be selected from $X_1$ or $X_2$ or $\dots$ or $X_t$ is $n_1 + n_2 + \dots + n_t$

Big Idea: Because the set family is pairwise disjoint, we can think of these sets like bubbles on a Venn Diagram that don't overlap. Counting total elements across these sets is the same as just adding the number of elements from each.

Example: A six person committee Alice, Ben, Connie, Dolph, Egbert, and Francisco is to select a chairperson, secretary, and treasurer.

In how many ways can this be done? $6\cdot5\cdot4=120$
In how many ways can this be done if Alice or Ben must be chairman? $5\cdot4 + 5\cdot 4 = 40$
How many ways if Egbert holds one of the offices? $\begin{align} 5\cdot4 + 5\cdot4 + 5\cdot4 &= 60\\ 3(5\cdot4) &= 60 \end{align}$
How many ways if both Dolph and Francisco holds one of the offices? $4\cdot3\cdot2=24$

Inclusion-Exclusion Principle for Two Sets: $|X \cup Y| = |X| + |Y| - |X \cap Y|$

Big Idea: If two sets intersect, adding the cardinality of both individually would double-count the overlap so it's subtracted away.

Permutation: A permutation of $n$ distinct elements $x_1, \dots, x_n$ is an ordering of the $n$ elements $x_1, \dots, x_n$ . There are $n!$ permutations of $n$ elements.

Example: ABCDEF Permutations

How many permutations of the letters ABCDEF are there with the substring DEF?
- Consider DEF to be a single element: $4! = 24$
How many permutations of the letters ABCDEF are there with DEF together in any order?
- Number of ways to order DEF is $3!$ . The number of permutations with DEF together is $3!4! = 144$
How many ways can you seat 6 people around a round table? (if everybody moves $n$ seats to the left or right, that is considered the same seating) $5!$

R-Permutation: An r-permutation of $n$ distinct elements $x_1, \dots, x_n$ is an ordering of an r-element subset of $x_1, \dots, x_n$ . The number of r-permutations of n distinct elements is $P(n, r) = \frac{n!}{(n-r)!}$

Example: How many ways can we select a chairperson, vice-chairperson, secretary, and treasurer froma group of 10 persons?

$P(10, 4) = 10 \cdot 9 \cdot 8 \cdot 7$

Example: In how many ways can seven distinct Martians and five distinct Neptunians wait in line if no two Neptunians stand together?

The Martians can stand together however they'd like, so there are $7!$ options for them. Because the Neptunians must have a Martian between them, there are 8 possible positions for them to stand. This means there are $P(8, 5)$ possible orderings of Neptunians.

$7! \cdot P(8, 5) = 33,868,800$

Combination: An r-combination of $X$ is an unordered selection of r elements of $X$ . The number of r-combinations of $n$ distinct objects is $C(n, r) = \frac{P(n, r)}{r!} = \frac{n!}{(n-r)!r!}$

Notation: $C(n, r) = \binom{n}{r}$

Example: In how many ways can we select a committee of two women and three men from a group of 5 distinct women and 6 distinct men?

$C(5, 2) \cdot C(6, 3) = 200$

Example: How many 8-bit strings contain exactly 4 ones?

$C(8, 4) = 70$

CS394 Lectures

Lecture 2