The course features many counterintuitive results, but that doesn't mean common sense is useless. Use it to sanity-check answers and keep work reasonable.
Prefer self-annotating expressions. Writing $\binom{52}{5}$ is better than computing the raw number — it communicates what is being counted. Simplify only when it's trivially reducible (e.g., $\frac{4}{2 \cdot 1} = 2$).
Re-reading your own work doesn't catch errors — you'll repeat the same mistake. Instead:
When a problem says "$n$ people" or "$k$ balls," immediately label them $1$ through $n$ (or $1$ through $k$). This is not stated in the problem but is extremely useful.
As far as probability and nature are concerned, objects behave as if they are distinguishable and labeled, even if they look identical to you. Treating "identical" objects as truly indistinguishable leads to counting errors.
Example: 10 green balls in a jar all look the same, but mentally labeling them 1–10 gives the correct count. Same for "6 robberies in 6 districts" — number the robberies 1–6 and the districts 1–6.
10 people split into a team of 4 and a team of 6:
Pick the team of 4; whoever's left is the team of 6. This also proves $\binom{n}{k} = \binom{n}{n-k}$ by counting the same thing two ways.
10 people split into two teams of 5:
The division by 2 corrects for overcounting — picking $\{1,2,3,4,5\}$ first and $\{6,7,8,9,10\}$ second is the same split as the reverse. This correction is needed only when the two groups are interchangeable (same size, no distinguishing label like "Team A" vs. "Team B").
The 2×2 sampling table (with/without replacement, order matters/doesn't) has three entries that follow directly from the multiplication rule. The fourth — order doesn't matter, with replacement — is the tricky one.
Pick $k$ times from a set of $n$ objects, with replacement, where order doesn't matter. How many outcomes?
| Case | Formula gives | Why it's correct |
|---|---|---|
| $k = 0$ | $\binom{n-1}{0} = 1$ | Picking nothing: one way (do nothing) |
| $k = 1$ | $\binom{n}{1} = n$ | Picking once: replacement and order are irrelevant |
| $n = 2$ | $\binom{k+1}{1} = k+1$ | Two boxes, $k$ dots total: box 1 can have $0, 1, \ldots, k$ dots |
The $n = 2$ case is the simplest non-trivial example — a generally useful research heuristic.
Choosing $k$ times from $n$ objects (with replacement, unordered) is equivalent to placing $k$ indistinguishable particles into $n$ distinguishable boxes.
Represent a configuration as a sequence of dots (particles) and vertical bars (separators between boxes).
Any valid configuration maps to exactly one such sequence and vice versa. The sequence always contains:
To specify a configuration, choose which $k$ of the $n + k - 1$ positions are dots. That gives $\binom{n+k-1}{k}$ ways. Equivalently, choose where the $n - 1$ separators go: $\binom{n+k-1}{n-1}$ — same number.
For two fair coins, the standard model has 4 equally likely outcomes: HH, HT, TH, TT. In 1925, Bose proposed (for particles, not coins) that indistinguishable objects yield only 3 outcomes — HH, TT, and "one of each" — all equally likely. Einstein supported the idea; together they predicted Bose-Einstein condensates, confirmed experimentally ~70 years later.
For probability problems using the naive definition: use the labeled (distinguishable) model. The Bose-Einstein model is relevant in quantum physics but not the default for counting or probability.
A story proof (proof by interpretation) establishes an identity by counting the same quantity two different ways, using a concrete combinatorial scenario. Contrast with algebraic proofs, which manipulate factorial expressions.
Advantages over algebra:
Story: Pick $k$ people from $n$ to include, or equivalently pick $n - k$ people to exclude. Same partition, same count.
Story: Choose a committee of $k$ from $n$ people, with one member designated president.
Both count the same thing, so they're equal.
Algebraic proof is difficult (factorials inside a sum). Story proof is clean:
Story: Pick $k$ people from a population of $m + n$, where the population consists of two groups of sizes $m$ and $n$.
If $j$ people come from the first group, then $k - j$ must come from the second. The number of ways to pick $j$ from the first group and $k - j$ from the second is $\binom{m}{j}\binom{n}{k-j}$. Summing over all valid splits $j = 0, 1, \ldots, k$ gives the total.
The naive definition requires (1) finitely many outcomes and (2) equally likely outcomes. The general definition removes both restrictions.
A probability space consists of $(S, P)$:
An event $A$ is a subset of $S$. We say $A$ occurs if the observed outcome $s_0 \in A$.
Only two rules are needed. Every theorem in probability theory follows from them.
$$P(\varnothing) = 0 \qquad P(S) = 1$$
Impossible events have probability zero; the certain event has probability one.
If $A_1, A_2, A_3, \ldots$ are disjoint (non-overlapping) events, then:
$$P\!\left(\bigcup_{i=1}^{\infty} A_i\right) = \sum_{i=1}^{\infty} P(A_i)$$
This extends to countably infinite unions, not just finite ones.