Chapter 5 of 5

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CAT Quantitative Ability★ ~3 of QA's 22 questions — CAT's smallest individual QA topic by weight, but highly template-driven

Modern Maths

Permutations, combinations, probability, special series and polygon-diagonal counting — a compact but fast-scoring topic once the recurring sub-type is recognised, since each sub-type has a fixed, learnable template.

⏱ 18 min readHard difficulty✓ Free · exam-pattern aligned

Timed practiceMCQs from this chapter with instant scoring

FlashcardsFormulas & traps on spaced repetition

CAT PYQsSolved previous-year questions

By the end of this chapter you'll be able to…

1Distinguish permutations from combinations, and apply the repeated-item and circular-arrangement formulas
2Apply the stars-and-bars formula for selection with unlimited repetition
3Solve 'at least one' counting and probability questions via the complement
4Compute probability for dice, cards (with and without replacement) and coin-toss settings
5Apply the special-series identities, including the sum of odd numbers and telescoping sums
6Count diagonals, lines and triangles from a polygon's vertices using combinations
7Compute a derangement count for a 'none in its original position' question

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Why this chapter matters in CAT

Permutations, combinations, probability, special-series sums and polygon-diagonal counting each reduce to a small set of recognisable sub-types, so the real skill is correctly identifying which sub-type a question belongs to — order-matters vs order-doesn't, with-replacement vs without, at-least-one vs exact-count — rather than the computation itself, which is then routine.

Before you start — revise these

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Algebra

Set-theory instincts (union, intersection) transfer directly into the probability addition and complement rules.

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Number System

Factorial manipulation and basic combinatorial identities recur throughout this chapter.

Modern Maths — CAT Quantitative Ability

Modern Maths is CAT's smallest individual QA topic by weight, but it is also the most template-driven: permutations and combinations, probability, special-series sums and polygon-diagonal counting each reduce to a small, fixed set of recognisable sub-types. The difficulty is almost always in correctly identifying which sub-type a question belongs to, not in the computation once it is identified.

1. Permutations and combinations

The fundamental principle of counting underlies every formula in this section: if one stage of a task can be done in $m$ ways and a second, independent stage in $n$ ways, the whole task can be done in $m \times n$ ways. Every permutation and combination formula is this principle applied to a specific structure, not a separate rule to memorise in isolation.

$n P r = \frac{n !}{( n - r )!} (order matters), n C r = \frac{n !}{r ! ( n - r )!} (order does not matter)$

Two identities make combinations faster to compute and to sanity-check. $n C r = n C (n - r)$ — choosing $r$ items to include is equivalent to choosing $n - r$ items to exclude, so $10 C 8$ is more easily computed as $10 C 2 = 45$ than by expanding $10 C 8$ directly. Pascal's identity, $n C r = (n - 1) C (r - 1) + (n - 1) C r$ , is the recursive rule behind Pascal's triangle and occasionally shortcuts a sum of several combination terms.

The single question that decides between the two is whether rearranging the same selected items counts as a different outcome. Choosing a 3-person committee from 10 people is a combination (the committee is the same regardless of the order named); arranging 3 of 10 books on a shelf is a permutation (a different order is a genuinely different arrangement).

Arrangements with repeated items divide out the repetitions: the number of distinct arrangements of a word with $n$ letters, where one letter repeats $p$ times, another $q$ times, and so on, is $\frac{n !}{p ! q ! \dots}$ . For the word MISSISSIPPI (11 letters: M once, I four times, S four times, P twice): $\frac{11 !}{4 ! 4 ! 2 !} = 34, 650$ distinct arrangements — treating every repeated letter as distinguishable would wildly overcount.

Circular arrangements of $n$ distinct people around a round table number $(n - 1)!$ , not $n!$ — fixing one person's seat removes the rotational symmetry that would otherwise count the same circular arrangement $n$ times (once for each rotation). If the arrangement's mirror image (clockwise vs anticlockwise) is also considered identical, as with a necklace of beads, the count is further halved to $\frac{( n - 1 )!}{2}$ .

Selecting items with repetition allowed (choosing $r$ items from $n$ types, unlimited supply of each, order irrelevant) uses the stars-and-bars formula:

$(r n + r - 1)$

Distributing 10 identical candies among 3 children (some may get zero) is $(3 - 1 10 + 3 - 1) = (2 12) = 66$ ways — a distinct formula from ordinary combinations, since here the "items" being chosen are unlimited-supply types, not distinct objects.

"At least one" counting questions are answered fastest through the complement, exactly as in Arithmetic's percentage chapter: count the arrangements or selections with NONE of the restricted property, and subtract from the total. "How many 3-person committees from 5 men and 4 women include at least one woman?" is faster as $(3 9) - (3 5)$ (total minus all-men) than by separately counting committees with exactly 1, 2 or 3 women and adding them.

Dividing $n$ distinct items into groups of specified sizes uses a direct extension of the arrangement-with-repetition idea: splitting 9 distinct people into groups of 4, 3 and 2 is $\frac{9 !}{4 ! 3 ! 2 !}$ when the groups themselves are distinguishable (e.g. named teams), but must be divided further by $k!$ if $k$ of the groups are of equal size and interchangeable (unnamed).

A derangement is a permutation in which no item ends up in its original position — the classic framing is $n$ letters placed into $n$ addressed envelopes with none in its correct envelope. The count is given by:

$D_{n} = n! \sum_{k = 0}^{n} \frac{( - 1 ) ^{k}}{k !}$

For $n = 4$ : $D_{4} = 4! (1 - 1 + \frac{1}{2} - \frac{1}{6} + \frac{1}{24}) = 9$ . A useful approximation for larger $n$ is $D_{n} \approx n! / e$ , which converges quickly — even at $n = 4$ it already gives $24/ e \approx 8.8$ , close to the exact value of 9.

2. Probability

$P (event) = \frac{favourable outcomes}{total outcomes} (equally likely outcomes only)$

Two dice, cards, and coins are the three recurring probability settings, and each has its own "total outcomes" trap. Two dice have 36 equally likely ordered outcomes, not 21 unordered sums — treating $(2, 5)$ and $(5, 2)$ as one outcome undercounts. A standard deck has 52 cards, and drawing without replacement changes the total for each subsequent draw, so the second card's probability is conditioned on what the first draw removed.

Conditional and sequential probability under without-replacement drawing is computed by multiplying stage-by-stage probabilities, each on the reduced remaining set — this is the same multiplication rule used for dependent events in Arithmetic's time-work chapter, applied here to a shrinking sample instead of a shrinking tank. For coins, the number of ways to get exactly $r$ heads in $n$ tosses is $(r n)$ out of $2^{n}$ total equally likely sequences — a direct application of combinations to a probability setting.

The addition rule, $P (A \cup B) = P (A) + P (B) - P (A \cap B)$ , and the complement rule, $P (at least one) = 1 - P (none)$ , both transfer directly from Algebra's set-theory instincts into probability. A question asking for the probability of drawing at least one ace in two draws (without replacement) is fastest via the complement: $1 - P (no aces) = 1 - \frac{48}{52} \times \frac{47}{51}$ , rather than separately summing the probabilities of exactly one ace and exactly two aces.

Two events are independent when the occurrence of one does not change the probability of the other — a fresh coin toss after a head is still a fresh 50-50 toss, unaffected by the previous result. This is the one place where "the coin is due for a tail" reasoning (the gambler's fallacy) is worth naming explicitly as a trap: independent trials genuinely have no memory of what came before.

3. Special series sums

$\sum_{k = 1}^{n} k = \frac{n ( n + 1 )}{2}, \sum_{k = 1}^{n} k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}, \sum_{k = 1}^{n} k^{3} = [\frac{n ( n + 1 )}{2}]^{2}$

The sum of the first $n$ odd numbers is exactly $n^{2}$ , a fact worth remembering directly rather than deriving from the general AP sum formula every time — $1 + 3 + 5 + \dots + 19$ (the first 10 odd numbers) is $1 0^{2} = 100$ with no term-by-term addition needed. The sum of the first $n$ even numbers is $n (n + 1)$ , exactly one AP-sum step away from the odd-number identity.

The sum-of-cubes identity has a striking structural feature worth noticing: it is literally the square of the sum-of-integers formula, $[\sum k]^{2} = \sum k^{3}$ — a fact that occasionally shortcuts a question asking to compare or relate the two sums directly, without computing either sum numerically first.

A fraction series whose terms telescope collapses almost entirely on subtraction, and recognising this structure turns an apparently long sum into a two-term computation. Writing each term $\frac{1}{k ( k + 1 )}$ as the difference $\frac{1}{k} - \frac{1}{k + 1}$ :

$\sum_{k = 1}^{n} \frac{1}{k ( k + 1 )} = (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \dots + (\frac{1}{n} - \frac{1}{n + 1}) = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}$

every interior term cancels against its neighbour, leaving only the first and last. Spotting that a series is telescoping — each term is a difference of consecutive values of some simpler expression — is far faster than attempting to sum the terms directly.

4. Polygon diagonal counting

The number of diagonals in a convex polygon with $n$ vertices is the total number of line segments between any two vertices, minus the $n$ sides that are not diagonals:

$Diagonals = (2 n) - n = \frac{n ( n - 3 )}{2}$

For a decagon ( $n = 10$ ): $(2 10) - 10 = 45 - 10 = 35$ diagonals. This is a direct application of combinations — choosing 2 vertices from $n$ gives every possible connecting segment, and subtracting the $n$ sides (which connect only adjacent vertices, not diagonally) leaves exactly the diagonals.

The same "choose 2 (or 3) vertices" idea answers two related counting questions that are easy to confuse with the diagonal count. The number of straight lines determined by $n$ points, no three of which are collinear, is simply $(2 n)$ — every pair of points determines one line, and this count includes the polygon's own sides, unlike the diagonal count, which explicitly excludes them.

The number of triangles formed by choosing any 3 of the $n$ vertices of a convex polygon is $(3 n)$ , since any 3 non-collinear points determine exactly one triangle regardless of whether the connecting segments are sides or diagonals of the polygon.

Worked Examples

Example 1 (permutations vs combinations — easy). In how many ways can a committee of 3 be chosen from 8 people, and in how many ways can 3 of those 8 be arranged in a line for a photo?

Committee (order irrelevant): $(3 8) = \frac{8 !}{3 ! 5 !} = 56$ . Photo arrangement (order relevant): $8 P 3 = \frac{8 !}{5 !} = 336$ — exactly $3! = 6$ times larger, since each committee of 3 can be arranged in $3!$ distinct orders.

Example 2 (arrangements with repetition — medium). Find the number of distinct arrangements of the letters of the word "BANANA."

BANANA has 6 letters: B once, A three times, N twice. Distinct arrangements $= \frac{6 !}{3 ! 2 !} = \frac{720}{12} = 60$ .

Example 3 (circular arrangement — medium). In how many ways can 6 people be seated around a round table?

$(6 - 1)! = 5! = 120$ ways, since fixing one person's seat removes the rotational duplicates.

Example 4 (stars and bars — hard). In how many ways can 10 identical candies be distributed among 3 children, with no restriction on how many each child receives (including zero)?

$(3 - 1 10 + 3 - 1) = (2 12) = 66$ ways.

Example 5 (probability, dice — easy). Two fair dice are rolled. Find the probability that the sum is 8.

Favourable outcomes (out of 36 ordered pairs): $(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)$ — 5 outcomes. Probability $= \frac{5}{36}$ .

Example 6 (probability, cards without replacement — hard). Two cards are drawn without replacement from a standard 52-card deck. Find the probability that both are aces.

$P = \frac{4}{52} \times \frac{3}{51} = \frac{12}{2652} = \frac{1}{221}$ — equivalently $\frac{( 2 4 )}{( 2 52 )} = \frac{6}{1326} = \frac{1}{221}$ , confirming the sequential and the combination approaches agree.

Example 7 (probability, coins — easy). Three fair coins are tossed. Find the probability of getting exactly 2 heads.

$\frac{( 2 3 )}{2 ^{3}} = \frac{3}{8}$ .

Example 8 (special series — easy). Find the sum of the first 10 odd natural numbers, and the sum of the squares of the first 10 natural numbers.

Sum of first 10 odd numbers $= 1 0^{2} = 100$ . Sum of squares $= \frac{10 \times 11 \times 21}{6} = 385$ .

Example 9 (polygon diagonals — medium). How many diagonals does a convex decagon (10 sides) have?

$(2 10) - 10 = 45 - 10 = 35$ diagonals.

Example 10 (at-least-one, complement counting — hard). From 5 men and 4 women, a committee of 3 is to be formed. In how many ways can the committee include at least one woman?

Total committees of 3 from 9 people: $(3 9) = 84$ . Committees with no woman (all men): $(3 5) = 10$ . At least one woman: $84 - 10 = 74$ .

Example 11 (polygon, triangles from vertices — medium). How many triangles can be formed by joining any 3 vertices of a convex octagon (8 vertices), and how does this compare with its diagonal count?

Triangles: $(3 8) = 56$ , since any 3 non-collinear vertices determine exactly one triangle. Diagonals: $(2 8) - 8 = 28 - 8 = 20$ . The two counts measure genuinely different things — one counts unordered vertex triples, the other counts non-adjacent vertex pairs — and it is a coincidence-free comparison, not a derivable relationship between the two.

Example 12 (special series, telescoping — hard). Find $k = 1 \sum 9 \frac{1}{k ( k + 1 )}$ .

Writing each term as $\frac{1}{k} - \frac{1}{k + 1}$ and summing, every interior term cancels: $1 - \frac{1}{10} = \frac{9}{10}$ .

Summary

Modern Maths is template-driven — the real skill is identifying which of a small number of recurring sub-types a question belongs to, since the computation itself is then routine.

Permutations count arrangements where order matters; combinations count selections where it does not. Repeated-item arrangements divide by the factorial of each repetition count; circular arrangements use $(n - 1)!$ , not $n!$ ; and selecting with unlimited repetition uses the stars-and-bars formula $(r n + r - 1)$ .

Dice, cards and coins are the three recurring probability settings — two dice give 36 ordered outcomes, card draws without replacement shrink the total at each stage, and coin-toss counts use combinations directly.

The sum of the first $n$ odd numbers is $n^{2}$ ; the sum of cubes is the square of the sum of integers. Polygon diagonals are $(2 n) - n$ , a direct combinations application once the $n$ sides are excluded from the vertex-pair count.

Key formulas & results

Everything to memorise for the exam hall, in one card. Screenshot this for revision.

Permutations and combinations

n P r = \frac{n !}{( n - r )!}, n C r = \frac{n !}{r ! ( n - r )!}

Order matters for permutations, not for combinations.

Combination identities

n C r = n C (n - r), n C r = (n - 1) C (r - 1) + (n - 1) C r

The first often simplifies computation; the second is Pascal's triangle's recursive rule.

Arrangement with repeated items

\frac{n !}{p ! q ! \dots}

Divides out the overcounting from each group of identical items.

Circular arrangement

(n - 1)! (or \frac{( n - 1 )!}{2} if mirror images are identical)

Fixing one seat removes the rotational duplicates.

Selection with repetition (stars and bars)

(r n + r - 1)

For distributing r identical items among n distinct recipients, any recipient possibly getting zero.

Complement / at-least-one

P (at least one) = 1 - P (none); count(at least one) = total - count(none)

Almost always faster than summing every 'exactly k' case separately.

Derangements

D_{n} = n! \sum_{k = 0}^{n} \frac{( - 1 ) ^{k}}{k !} \approx \frac{n !}{e}

Counts permutations with no item in its original position.

Special series

\sum k = \frac{n ( n + 1 )}{2}, \sum k^{2} = \frac{n ( n + 1 ) ( 2 n + 1 )}{6}, \sum k^{3} = [\frac{n ( n + 1 )}{2}]^{2}, \sum (first n odd) = n^{2}

The sum of cubes is literally the square of the sum of integers.

Telescoping series

\sum_{k = 1}^{n} \frac{1}{k ( k + 1 )} = 1 - \frac{1}{n + 1} = \frac{n}{n + 1}

Each term splits into a difference of consecutive values; interior terms cancel.

Polygon counting

Diagonals = (2 n) - n, Lines = (2 n), Triangles = (3 n)

All three are direct combinations applications on the n vertices.

CAT test series

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Modern Maths shows up in every CAT paper — sit a timed one and find out where you actually stand.

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Traps CAT sets — and how to dodge them

These are the exact option-traps and misreads that cost marks under negative marking.

WATCH OUT

✗ Using a permutation formula when the question describes an unordered selection

✓ Ask first whether rearranging the same chosen items would count as a different outcome; use combinations if not.

Why it happens: Committees, teams and unordered selections are combinations; arrangements, rankings and seatings are permutations.

WATCH OUT

✗ Using n! for a circular arrangement instead of (n-1)!

✓ Fix one person's position to remove the rotational symmetry, giving (n-1)! distinct arrangements.

Why it happens: n! treats each rotation of the same circular arrangement as a different outcome, which overcounts by a factor of n.

WATCH OUT

✗ Forgetting to divide by the factorial of each repeated letter/item count

✓ Use n!/(p!q!...) for arrangements with repeated items.

Why it happens: Treating repeated items as distinguishable overcounts every arrangement by the number of ways the repeated items could be permuted among themselves.

WATCH OUT

✗ Computing 'at least one' by summing every exact-count case

✓ Compute the complement (none) and subtract from the total instead.

Why it happens: Summing exact cases is more error-prone and slower than one complement subtraction, especially as the number of cases grows.

WATCH OUT

✗ Treating two dice as having 21 unordered sum-outcomes

✓ Use all 36 equally likely ordered outcomes as the sample space.

Why it happens: Unordered sums are not equally likely — a sum of 7 has 6 ordered ways to occur while a sum of 2 has only 1, so treating sums as equally likely outcomes gives wrong probabilities.

WATCH OUT

✗ Using the same probability for a second card draw as the first, ignoring the first draw's effect

✓ Recompute the total and favourable counts for each subsequent draw under without-replacement conditions.

Why it happens: Removing a card changes both the deck size and the count of the favourable outcome for every later draw.

WATCH OUT

✗ Reasoning that a coin is 'due' for an outcome after a run of the opposite result

✓ Independent trials have no memory; each toss remains a fresh 50-50 probability regardless of history.

Why it happens: This is the gambler's fallacy — treating independent events as if they compensate for past outcomes.

Exam-pattern practice

PYQ-style questions with full solutions. Work through them as a readiness check — mark yourself honestly and get your gap report at the end.

Readiness check

Are you exam-ready for Modern Maths?

12 problems from this chapter. Try each one, reveal the worked solution, mark yourself honestly — get your gap report at the end.

12 questions~8 min worth ~66 marks in CAT exams

5-minute revision

The whole chapter, distilled. Read this the night before the exam.

•Order matters -> permutations; order doesn't -> combinations.
•nCr = nC(n-r); Pascal's identity nCr = (n-1)C(r-1) + (n-1)Cr.
•Repeated-item arrangements divide by the factorial of each repetition count.
•Circular arrangements use (n-1)!, halved further if mirror images are identical.
•Selection with unlimited repetition uses stars-and-bars: C(n+r-1, r).
•'At least one' is almost always fastest via the complement: total minus none.
•Two dice have 36 ordered outcomes, not 21 unordered sums.
•Without-replacement draws shrink both the total and the favourable count at each stage.
•Independent trials have no memory — the gambler's fallacy is a trap, not a strategy.
•Sum of first n odd numbers is n²; sum of cubes is the square of the sum of integers.
•A telescoping series collapses to its first and last terms once each term is split into a difference.
•Polygon diagonals, lines and triangles from vertices are all direct combinations applications.
•Derangements: D_n = n!·sum of (-1)^k/k!, approximated well by n!/e.

CAT question blueprint

How this topic is asked, tier by tier — so you can prep to the pattern.

Typical weightage: Modern Maths contributes an estimated 6-9 of QA's 66 marks (about 3 of 22 questions)

Question style	Marks each	Typical count	What it tests
Permutations and combinations	3	~1	Basic P&C, repeated-item arrangements, circular arrangements and stars-and-bars
Probability — dice, cards, coins	3	~1	Classical probability, without-replacement sequencing and complement counting
Special series and sequences	3	~0-1	Sum-of-odd-numbers, sum of squares/cubes and telescoping series
Polygon counting	3	~0-1	Diagonals, lines and triangles from a polygon's vertices
Complement / at-least-one counting	3	~0-1	At-least-one selection and probability questions
Derangements	3	~0-1	Counting permutations with no fixed points
Circular arrangements	3	~0-1	Seating around a round table, with or without mirror-image symmetry
Selection with repetition	3	~0-1	Stars-and-bars distribution counting

Prep strategy

Day 1: fundamental counting principle, permutations vs combinations, and the two combination identities.
Day 2: repeated-item arrangements, circular arrangements, and stars-and-bars.
Day 3: probability — dice, cards (with and without replacement), coins, and the complement rule.
Day 4: special series (including telescoping), polygon counting, and derangements, then a timed mixed set.

Exam-hall strategy

Battle-tested tips from mentors and toppers for this topic under the sectional clock.

Identify order-matters vs order-doesn't before choosing a permutation or combination formula.
For 'at least one' questions, compute the complement rather than summing exact cases.
For sequential without-replacement draws, recompute the total and favourable counts at every stage.
Check whether a series' terms can be rewritten as a telescoping difference before summing term by term.
Use the combinations-based diagonal/line/triangle formulas directly for any polygon-vertex counting question.

Beyond the exam

Where this skill shows up in the job you're competing for — and in life.

Scheduling and resource allocation

Combinations and stars-and-bars directly model how many ways limited resources or slots can be assigned among people or tasks.

Quality control and risk estimation

Probability without replacement is the exact model behind sampling inspection and defect-rate estimation from a finite batch.

Cryptographic key spaces

Permutation and arrangement counts underlie the size of possible key spaces in basic cryptographic and combinatorial security arguments.

Where else this topic is tested

Prepare once, score in every exam that asks it.

XAT Quantitative Ability & DIHigh — near-identical Modern Maths topics

SSC CGL / IBPS PO Quantitative AptitudeModerate — simpler permutation-combination and probability applications

GATE Discrete MathematicsModerate overlap on combinatorics and probability, at a more formal level

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Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Roughly 3 of QA's 22 questions — the smallest individual QA topic by count, but highly scorable once the recurring sub-type (permutation vs combination, with vs without replacement, at-least-one vs exact) is correctly identified, since the computation itself is then a fixed, learnable template.

Ask whether rearranging the same selected items would count as a genuinely different outcome. If yes (seating, ranking, arranging on a shelf), it's a permutation. If no (forming a committee, choosing a team), it's a combination.

Because rotating everyone around the table by one seat produces the same circular arrangement, not a new one, and n! would count each of the n rotations of the same arrangement as distinct. Fixing one person's seat removes this duplication entirely.

Compute the complement — the count or probability of NONE having the property — and subtract from the total or from 1. This is almost always faster and less error-prone than summing every 'exactly k' case.

A series whose terms can each be rewritten as a difference of consecutive values of some simpler expression, so that all interior terms cancel when the series is summed, leaving only the first and last. A term like 1/(k(k+1)), rewritten as 1/k − 1/(k+1), is the classic recognisable shape.

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Modern Maths

By the end of this chapter you'll be able to…

Before you start — revise these

Modern Maths — CAT Quantitative Ability

1. Permutations and combinations

2. Probability

3. Special series sums

4. Polygon diagonal counting

Worked Examples

Summary

Key formulas & results

Traps CAT sets — and how to dodge them