🔢 Quant

Sets, Counting & Probability

The two-set identity, permutations vs combinations, complement counting, and probability with and without replacement.

~6h

to master

Sets: the two-circle identity

For any finite sets, $|A \cup B| = |A| + |B| - |A \cap B|$ — adding the circles double-counts the overlap, so subtract it once. With "each of 25 students takes history or math (or both), 20 take history, 18 take math": $25 = 20 + 18 - x$ gives $x = 13$ in both.

Disjoint sets: overlap $= 0$ , so sizes simply add.
"Neither" lives outside both circles: total $= |A \cup B| + \text{neither}$ — forgetting it is the classic miss.
Three circles exist on the GMAT but are rare; a two-way table is usually faster anyway.

Counting: multiply choices, then correct for order

📐Core Rule

Multiplication principle: independent choices multiply. 5 entrées × 3 desserts = 15 meals; 8 coin flips = $2^8$ outcomes.

From it, everything else follows:

Factorials count orderings: $n$ distinct objects line up in $n! = n(n-1)\cdots 1$ ways. ( $0! = 1! = 1$ .)
Permutations (order matters): filling $k$ slots from $n$ options = $n(n-1)\cdots(n-k+1)$ .
Combinations (order doesn't): $\binom{n}{k} = \dfrac{n!}{k!(n-k)!}$ — divide out the $k!$ orderings of each chosen group. $\binom{5}{2} = 10$ , and the symmetry $\binom{n}{k} = \binom{n}{n-k}$ saves arithmetic: choosing 2 to include is choosing 3 to leave out.

💡Exam Tip

"Order matters?" is the only diagnostic you need. Committees, handshakes, pairings → combinations. Rankings, seatings, codes → permutations/slots. When unsure, ask: does swapping two selections create a new outcome?

Repeated letters deflate counts: arrangements of

n

objects where one letter repeats

r

times

= \frac{n!}{r!}

. Constraint counting ("the two I's must not be adjacent") is usually fastest by complement: count everything, subtract the violators — glue the pair into one block to count them. That exact move solves .

Probability: count two things

For equally likely outcomes, $P(E) = \dfrac{\text{favourable}}{\text{total}}$ — so probability is counting, twice. Always $0 \le P \le 1$ .

📐Core Rule

The combination rules: $P(\text{not } E) = 1 - P(E)$ — the complement is often 90% of the work saved ("at least one" → $1 - P(\text{none})$ ). $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ — the set identity in probability clothes. Independent events: $P(A \text{ and } B) = P(A) \cdot P(B)$ . Three independent solvers succeed/fail per their own rates: P(X and Y succeed, Z fails) $= p_X \, p_Y (1 - p_Z)$ .

Drawing without replacement updates the pool: after two non-stock cards leave a 48-card deck holding 8 stock cards, the third draw's chance is

\frac{8}{46} = \frac{4}{23}

— recount numerator and denominator after every draw. That's .

⚠️GMAT Trap

"Or" double-counts unless you subtract the overlap; "and" multiplies only under independence. Rolling one die, P(odd or prime) $= \frac{3}{6} + \frac{3}{6} - \frac{2}{6} = \frac{4}{6}$ — the naive sum gives 1, absurd. And successive draws from one deck are not independent.

⚡Shortcut

For multi-stage probabilities, multiply along the branch: P(first ace, then king) $= \frac{4}{52} \cdot \frac{4}{51}$ . Writing the branch out beats formula-hunting every time.

Checklist

Union problems → two-set identity; park "neither" outside
Decide order-matters before picking a formula
Repeats → divide by $r!$ ; adjacency bans → complement + glue
"At least one" → complement
Without replacement → re-count after each draw

Sample Questions

22 practice questions

Hard

A box contains 30 batteries, exactly 6 of which are defective. Two are drawn at random without replacement. What is the probability that BOTH are defective?

Hard

Using each letter of LEVEL exactly once, how many distinct 5-letter arrangements can be formed in which the two L's are NOT next to each other?

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