The Timpson Lyceum · Discrete Math (2025)

Lists – Lecture 4 (Discrete Math)

Lists – Lecture 4

Overview

This lecture introduces lists as the first kind of collection on the way to a broader study of counting and combinatorial proof. A list is an ordered sequence of objects, where both order and repetition matter — a feature that distinguishes lists from sets. After defining lists and the special case of ordered pairs, the lecture develops the core counting rules for how many lists of a given length can be formed from a fixed pool of options, both with and without repeats. These rules — and the falling-factorial notation $n_k$ — are the two formulas that drive the chapter's counting problems.

Key Concepts

List. An ordered collection of objects, written with parentheses and comma-separated entries, e.g. $(1, 2, 3)$. An "object" is undefined — it can be anything, including a set or even another list (e.g. $(\text{SpongeBob}, \{1,2,3\}, 2, 1)$ is a list of four objects, and $((1), (1,2))$ is a list of two objects, each itself a list).

Equality of lists. Two lists are equal if and only if they have the same length and agree entry-by-entry: $(a, b, c) = (x, y, z) \iff a = x,\ b = y,\ c = z.$ This generalizes to lists of any length: the $i$-th entries must match for every $i$.

• Order matters: $(1, 2, 3) \neq (3, 2, 1)$.
• Duplicates matter: $(1, 2, 3) \neq (1, 1, 2, 3)$.

Both properties separate lists from sets, where $\{1,2,3\} = \{3,2,1\}$ and repeats are ignored.

Ordered pair. A two-element list. These are everywhere: the coordinate $(3, 4)$ in the $xy$-plane is an ordered pair used to signify a point. The point is a geometric object; the pair is just a list. Because order matters, $(3,4)$ names a different location than $(4,3)$.

Counting lists — the two master rules. Let $n$ be the number of options available for each position and $k$ the length of the list.

• With repeats (replacement allowed): there are $n^k$ such $k$-element lists. Reasoning: each of the $k$ positions independently has $n$ choices, giving $n \cdot n \cdots n$ ($k$ times). For ordered pairs ($k=2$) this is $n^2$; e.g. ordered pairs from $\{1,2,3,4\}$ number $4^2 = 16$.

• Without repeats (no replacement): there are $n_k = n(n-1)(n-2)\cdots(n-k+1)$ such $k$-element lists ($k$ falling factors). This is the falling-factorial notation $n_k$: start at $n$ and multiply $k$ decreasing terms. The last factor is always $n - k + 1$.

Why the last factor is $n - k + 1$. For $10_4$: start at $10$ and take four factors — $10 \cdot 9 \cdot 8 \cdot 7$ — and $7 = 10 - 4 + 1$. Similarly $20_{10} = 20 \cdot 19 \cdots 11$.

Connection to factorial. $n! = n_n$, the falling factorial taken all the way down to $1$.

Two pools of different size (product rule). If the first position has $n$ options and the second has $m$ options, the number of ordered pairs is $n \cdot m.$ For example, first entry from $\{1,2,3\}$ and second from $\{a,b,c,d,e\}$ gives $3 \cdot 5 = 15$ pairs.

Combinatorial proof (product rule, $n \times m$). A representative argument: construct a chart in which each row collects all the pairs sharing a fixed first element. There is one row per first-element choice, so there are $n$ rows. For any fixed first element there are $m$ ways to choose the second, so each row holds $m$ pairs. Thus $n$ rows of $m$ pairs give $\underbrace{m + m + \cdots + m}_{n} = n \cdot m$ pairs total. Such counting proofs are necessarily wordier and less formally rigorous than algebraic proofs (like "even + even = even") — they argue from a clearly described construction rather than pure symbol manipulation.

Two ways to count "no repeats," same answer. Counting ordered pairs from $\{1,2,3,4\}$ with no repeated entry can be seen as either removing the $n$ "diagonal" pairs from the full grid, $n^2 - n$, or as $n$ rows each shortened to $n-1$ entries, $n(n-1)$. These agree: $n^2 - n = n(n-1)$. Different intuitions, identical result — there is no single "right" way to count.

Worked Example

Problem. A club of 10 people elects a president, vice-president, secretary, and treasurer (no one holds two offices). How many ways?

This is a 4-element list with no repeats drawn from $n = 10$ options, so apply the falling-factorial rule $n_k = n(n-1)\cdots(n-k+1)$ with $n = 10$, $k = 4$:

Each office is filled in turn — 10 choices for president, then 9 remaining for vice-president, 8 for secretary, 7 for treasurer — and the last factor is $7 = 10 - 4 + 1$, exactly as the notation predicts. (On a test, leaving the answer as $10_4$ or $10\cdot 9\cdot 8\cdot 7$ is preferred over the multiplied-out number, since it shows the counting reasoning.)

A useful sanity check on the formula's edges: selecting 5 objects from a bag of 3 without replacement is a 5-element list from $n=3$, i.e. $3_5 = 3 \cdot 2 \cdot 1 \cdot 0 \cdot (-1) = 0$ — correctly reporting that it cannot be done, since the third factor hits $0$.

Summary

• A list is an ordered collection of objects; order and duplicates both matter, so $(1,2,3) \neq (3,2,1) \neq (1,1,2,3)$.
• Lists are equal exactly when they match length and agree entry-by-entry; an ordered pair is just a two-element list (like an $xy$-coordinate).
• Number of $k$-element lists from $n$ options with repeats: $n^k$.
• Number of $k$-element lists from $n$ options without repeats: the falling factorial $n_k = n(n-1)\cdots(n-k+1)$, whose last factor is $n-k+1$; note $n! = n_n$.
• With two positions drawing from pools of size $n$ and $m$, the product rule gives $n \cdot m$ ordered pairs, justified by a chart of $n$ rows with $m$ entries each.
• The same count often has several valid derivations (e.g. $n^2 - n = n(n-1)$); the formulas even stay self-consistent in impossible cases, returning $0$.