Modern Math — XAT Quantitative Ability & Data Interpretation

Every sub-topic in this chapter has exactly one question that decides everything else: does order matter (permutation) or not (combination)? Is an element replaced before the next pick, or not? Did you double-count an overlap? Get that one question right first, and the rest is direct formula application.

1. What XAT actually asks

Modern Math is estimated at roughly 5-7 of QA&DI's 28 questions across recent papers (see docs/exam-briefs/xat-2026-brief.md), covering permutations and combinations, probability, arithmetic/geometric progressions, and set theory. These sub-topics are grouped together because they share a common character: each rests on a small number of clean formulas, and the entire difficulty is correctly identifying which formula the question's wording actually calls for.

2. Permutations and combinations

Permutation (order matters): ${}^n{P}_r=\frac{n!}{(n-r)!}$ — arranging $r$ items from $n$ distinct items where the sequence matters (rankings, passwords, seating in distinct chairs).

Combination (order doesn't matter): ${}^n{C}_r=\frac{n!}{r!(n-r)!}$ — selecting $r$ items from $n$ where only the group composition matters (choosing a committee, picking a team).

The one question that decides everything: does swapping two selected items create a different outcome? If yes — permutation. If the outcome is the same selection either way — combination. "Arrange," "rank," "form a code" signal permutation; "select," "choose," "form a committee" signal combination.

Circular permutations of $n$ distinct items: $(n-1)!$, not $n!$ — because rotating the whole arrangement doesn't create a new one, one position can always be fixed as a reference.

3. Probability

$P(\text{event})=\frac{\text{number of favourable outcomes}}{\text{total number of possible outcomes}}$

Mutually exclusive events (cannot both happen): $P(A\text{ or }B)=P(A)+P(B)$. Independent events (one doesn't affect the other): $P(A\text{ and }B)=P(A)\times P(B)$.

With replacement vs. without replacement is this sub-topic's version of the permutation-vs-combination decision: drawing a card, replacing it, then drawing again keeps the total outcome count constant across draws; drawing without replacement shrinks the total by one each time, and probabilities for subsequent draws must be recalculated on the smaller remaining set.

Complementary probability is often the fastest path to an answer: $P(\text{at least one})=1-P(\text{none})$ — computing "none" is almost always easier than directly summing every "at least one" case.

4. Progressions

Arithmetic Progression (AP): consecutive terms differ by a constant $d$.

$n^{\text{th}}\text{ term}=a+(n-1)d,\text{Sum of }n\text{ terms}=\frac{n}{2}\left[2a+(n-1)d\right]$

Geometric Progression (GP): consecutive terms have a constant ratio $r$.

$n^{\text{th}}\text{ term}=a{r}^{n-1},\text{Sum of }n\text{ terms}=\frac{a(r^n-1)}{r-1}(r\ne 1)$

The tell between AP and GP: check whether consecutive differences are constant (AP) or consecutive ratios are constant (GP) — a sequence like 3, 6, 12, 24 has ratio 2 throughout (GP), while 3, 6, 9, 12 has difference 3 throughout (AP). Testing both quickly on the first three given terms resolves which formula applies before any further work.

5. Set theory

For two sets, the union-intersection rule avoids double-counting elements present in both:

$n(A\cup B)=n(A)+n(B)-n(A\cap B)$

For three sets, the extended version accounts for every possible overlap:

$n(A\cup B\cup C)=n(A)+n(B)+n(C)-n(A\cap B)-n(B\cap C)-n(A\cap C)+n(A\cap B\cap C)$

The recurring XAT question type ("X% failed subject A, Y% failed subject B, Z% failed both — what % failed neither") is a direct union-intersection application: find the union (failed at least one), then subtract from 100% to find the complement (failed neither).

7. Common traps

• Confusing permutation and combination — always ask first whether reordering the same selected items changes the outcome.
• Forgetting to adjust probabilities for "without replacement" draws — the denominator (and sometimes numerator) shrinks with each draw; squaring or cubing a single-draw probability is only valid with replacement.
• Missing the constant-ratio vs. constant-difference test before assuming a sequence is AP when it's actually GP, or vice versa.
• Double-counting the intersection in set theory "union" questions — always subtract the overlap once; for three sets, add back the triple overlap after subtracting all three pairwise overlaps.
• Computing "at least one" directly instead of via the complement — this is almost always slower and more error-prone than computing $1-P(\text{none})$.
• Using $n!$ instead of $(n-1)!$ for circular permutations — forgetting that rotating a circular arrangement doesn't create a genuinely new one.

8. When to guess, and why

Within your first 8 skips across all of Part 1, a modern math question involving a longer, multi-step calculation (a three-set union, a multi-draw probability chain) is a reasonable one to skip if a clean setup doesn't emerge quickly, since these sub-topics are especially prone to a single early misstep compounding through several later steps.

Beyond your 8th skip, a blind 1-in-5 guess has an expected value of exactly 0, while a blank costs -0.10 — mark your best remaining guess.

If you've correctly identified the sub-topic and formula but stalled on execution, a rough estimate (checking whether an answer's magnitude is even plausible, or whether it satisfies an obvious parity or divisibility check) often eliminates two or three options before guessing.

Summary

• Modern Math is roughly 5-7 of QA&DI's 28 questions, covering permutations/combinations, probability, progressions, and set theory.
• Permutation (order matters) vs. combination (order doesn't) is the single decision that determines every question in that sub-topic — check it first.
• With-replacement keeps draw probabilities constant across draws; without-replacement requires recalculating on a shrinking remaining set.
• Complementary probability ($1-P(\text{none})$) is almost always faster than directly computing "at least one."
• AP has a constant difference between consecutive terms; GP has a constant ratio — test both quickly on the first few given terms.
• The union-intersection formula prevents double-counting overlaps; the three-set version needs the triple overlap added back after all pairwise overlaps are subtracted.
• Circular permutations of $n$ items use $(n-1)!$, not $n!$, since rotation doesn't create a new arrangement.
• Past your 8th free skip in Part 1, a rough magnitude or parity check on a stalled multi-step calculation often narrows the field before a final guess is needed.