Chapter 4 of 16

25%

JEE Advanced Mathematics★ ~2 questions across Paper 1 and Paper 2 combined, usually one restricted-arrangement or distribution question and one that combines counting with probability

Permutations and Combinations

Why reserving a place over-counts, selections from a multiset, the four kinds of distribution, derangements and the rank of a word.

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

Timed practiceMCQs from this chapter with instant scoring

FlashcardsFormulas & traps on spaced repetition

JEE Advanced PYQsSolved previous-year questions

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

1Recognise when a counting argument double-counts, and repair it by splitting into disjoint cases or by counting the complement
2Count arrangements of objects that are not all distinct, and handle separation conditions using the gap method
3Count selections from a multiset by classifying them according to their shape rather than by a single binomial coefficient
4Distinguish the four distribution problems, and apply stars and bars, including with upper bounds via inclusion and exclusion
5Decide whether a grouping question requires division by the number of equal-sized groups
6Compute derangements, the rank of a word in dictionary order, and counts of geometric and digit configurations

💡

Why this chapter matters in JEE Advanced

Almost every mark lost in this chapter is lost to double counting rather than to arithmetic, and Advanced sets its questions to exploit exactly that. The habit of reserving places to satisfy a condition and then filling the rest freely produces a plausible number that is simply wrong, and the discipline of splitting into disjoint cases is what separates a correct solution from a confident one. The chapter also supplies the entire counting apparatus for Probability, so an error here propagates into a second question elsewhere on the paper.

Before you start — revise these

🔗

The addition and multiplication principles of counting

🔗

Factorials and the definitions of the permutation and combination symbols

🔗

Inclusion and exclusion for two and three sets

🔗

Divisibility tests for two, four and five

Permutations and Combinations

A committee of $5$ is chosen from $6$ men and $4$ women, and must contain at least $2$ women. A common line of reasoning says: pick $2$ women to satisfy the condition, then pick any $3$ of the remaining $8$ people. That gives $(2 4) (3 8) = 6 \times 56 = 336$ .

The true answer is $186$ .

The method double-counts. Take the committee ${W_{1}, W_{2}, W_{3}, M_{1}, M_{2}}$ . It is produced when the reserved pair is ${W_{1}, W_{2}}$ and $W_{3}$ arrives later, and again when the reserved pair is ${W_{1}, W_{3}}$ , and again when it is ${W_{2}, W_{3}}$ — three times over, and there is nothing in the count that removes the duplicates.

Two repairs work. Split into disjoint cases by the exact number of women:

$(2 4) (3 6) + (3 4) (2 6) + (4 4) (1 6) = 120 + 60 + 6 = 186.$

Or count the complement: from $(5 10) = 252$ total committees, remove those with no woman, $(5 6) = 6$ , and those with exactly one, $(1 4) (4 6) = 60$ , leaving $252 - 66 = 186$ .

This is the whole discipline of the chapter. Every valid count is a sum over cases that are disjoint and exhaustive, and the fastest way to get a wrong answer is to satisfy a condition first and then count freely, because the object no longer remembers which part satisfied the condition.

1. Arrangements when some objects are alike

Arranging $n$ objects of which $n_{1}$ are of one kind, $n_{2}$ of another and so on gives

$\frac{n !}{n _{1} ! n _{2} ! \dots n _{k} !},$

because each genuinely different arrangement is produced $n_{1}! n_{2}! \dots$ times when the identical objects are treated as distinct. Dividing by the size of the group that shuffles identical items is the same repair as in the opening, applied in the other direction.

Illustration 1

How many arrangements are there of the letters of MISSISSIPPI?

The letters are $I^{4}$ , $S^{4}$ , $P^{2}$ , $M$ , eleven in all, so the count is $\frac{11 !}{4 ! 4 ! 2 !} = 34650$ .

Illustration 2

In how many of those arrangements are no two $S$ s adjacent?

Arrange the seven non- $S$ letters first: $\frac{7 !}{4 ! 2 !} = 105$ ways. Those seven letters create eight gaps, including the two ends, and choosing four of them for the $S$ s guarantees no two are together: $(4 8) = 70$ . The answer is $105 \times 70 = 7350$ .

The gap method is the standard device for every "no two together" condition, and it works because placing the restricted objects into gaps makes the separation automatic rather than something to be checked afterwards.

2. Selecting from a multiset

Selecting $r$ objects from a collection containing repeats is not a single binomial coefficient, because the repeats limit how many copies of a letter a selection can hold. The reliable method is to split by the shape of the selection: how many pairs, how many singletons.

Illustration 3

From the letters of EXAMINATION, how many selections of four letters are possible, and how many four-letter words?

The word has eleven letters, of which $A$ , $I$ and $N$ appear twice each and $E, X, M, T, O$ once each, so there are eight distinct letters.

shape	selections	arrangements each	words
four different	$(4 8) = 70$	$4! = 24$	$1680$
one pair, two different	$3 \times (2 7) = 63$	$\frac{4 !}{2 !} = 12$	$756$
two pairs	$(2 3) = 3$	$\frac{4 !}{2 ! 2 !} = 6$	$18$
total	$136$		$2454$

The shapes are disjoint and exhaustive, which is what makes the addition legitimate. Notice that the arrangement count differs from shape to shape, so the selections cannot be totalled first and multiplied by a single factorial.

3. Distributions: four different questions

"Put $n$ objects into $r$ boxes" hides four questions, and the answer depends entirely on whether the objects and the boxes are distinguishable.

The identical-objects, distinct-boxes case is the one with a usable formula. Represent a distribution of $n$ identical objects among $r$ boxes as a row of $n$ stars separated by $r - 1$ bars; every arrangement of the $n + r - 1$ symbols is one distribution, so the count is $(r - 1 n + r - 1)$ . Requiring every box to be non-empty means placing one object in each first, leaving $(r - 1 n - 1)$ .

Illustration 4

Find the number of non-negative integer solutions of $x + y + z = 10$ , and the number with every variable at most $4$ .

Without restriction, $(2 12) = 66$ . For the upper bound, subtract the solutions in which some variable is at least $5$ : substituting $x^{'} = x - 5$ turns that condition into $x^{'} + y + z = 5$ , giving $(2 7) = 21$ each, so $63$ in all. Solutions where two variables are both at least $5$ have been removed twice; each gives $x^{'} + y^{'} + z = 0$ , one solution, and there are $3$ such pairs. So the count is

$66 - 63 + 3 = 6.$

Listing confirms it: the solutions are the permutations of $(4, 4, 2)$ and of $(4, 3, 3)$ , three each.

Read the other way, the same formula answers a selection question. Choosing $r$ objects from $n$ types with unlimited repetition, order irrelevant, is the same as deciding how many of each type to take, so the count is $(r n + r - 1)$ . Buying ten chocolates from four flavours, for instance, can be done in $(3 13) = 286$ ways. Selections with repetition and distributions of identical objects are one problem wearing two descriptions.

Illustration 5

In how many ways can $5$ distinct balls go into $3$ distinct boxes with no box empty?

This is exactly the count of onto functions, so inclusion and exclusion gives $3^{5} - 3 \cdot 2^{5} + 3 \cdot 1^{5} = 243 - 96 + 3 = 150$ . If instead the boxes were identical, we would divide by the $3!$ orderings of the boxes, giving $25$ .

Illustration 6

In how many ways can $5$ identical balls go into $3$ identical boxes?

Now nothing distinguishes either the balls or the boxes, so the answer is the number of ways to write $5$ as a sum of at most three positive parts, order irrelevant: $5$ , $4 + 1$ , $3 + 2$ , $3 + 1 + 1$ , $2 + 2 + 1$ . Five ways. There is no formula, and there is not meant to be — questions of this type are always small enough to list.

4. Division into groups, and when to divide by the number of groups

Splitting people into groups looks like a distribution problem, and it is — but with one extra decision that decides the answer. If the groups are labelled, say by the project each will work on, the multinomial coefficient is the whole answer. If the groups are merely a partition into unlabelled sets of equal size, the multinomial has counted each partition once for every way of ordering the groups, and that ordering must be divided out.

Dividing $n$ objects into groups of sizes $n_{1}, n_{2}, \dots, n_{k}$ that are all different gives $\frac{n !}{n _{1} ! n _{2} ! \dots n _{k} !}$ and needs no further division, because the sizes themselves tell the groups apart. When $m$ of the groups share the same size, divide by $m!$ .

Illustration 7

In how many ways can $9$ students be divided into three groups of three? And in how many ways can they be assigned to three distinct projects, three to each?

The multinomial count is $\frac{9 !}{3 ! 3 ! 3 !} = \frac{362880}{216} = 1680$ , and that is the answer for the labelled version, since a project name distinguishes the groups.

For unlabelled groups the same partition has been counted once for each of the $3!$ orders in which the three blocks could have been produced, so the answer is $\frac{1680}{6} = 280$ .

The two numbers differ by a factor of six, and choosing between them is decided by a single reading of the question: if swapping two entire groups produces a different outcome in the situation described, do not divide.

5. Circular arrangements

Seating $n$ people around a table produces the same arrangement $n$ times over, once for each rotation, so the count is $\frac{n !}{n} = (n - 1)!$ . If the arrangement can also be flipped over, as with a necklace of beads or a garland of flowers, each arrangement is counted twice more and the answer is $\frac{( n - 1 )!}{2}$ .

The distinction is physical, not verbal: a garland can be turned over and a dinner table cannot, so a question that says "necklace" or "garland" is asking for the halved count.

Illustration 8

Five men and five women are to be seated alternately around a round table. In how many ways?

Seat the men first: $(5 - 1)! = 24$ circular arrangements. They create five gaps, and the women fill them in $5! = 120$ ways. The total is $24 \times 120 = 2880$ .

Seating the men first is not arbitrary — fixing one group is what removes the rotational freedom, and doing it twice would remove it twice.

6. Derangements

A derangement is a permutation that moves every object. Counting them uses inclusion and exclusion over the positions that stay fixed:

$D_{n} = n! (1 - \frac{1}{1 !} + \frac{1}{2 !} - \dots + \frac{( - 1 ) ^{n}}{n !}),$

giving $D_{1} = 0$ , $D_{2} = 1$ , $D_{3} = 2$ , $D_{4} = 9$ , $D_{5} = 44$ . The recursion $D_{n} = (n - 1) (D_{n - 1} + D_{n - 2})$ is quicker for small $n$ and worth carrying.

Illustration 9

Five letters are placed at random into five addressed envelopes. In how many ways does exactly one letter reach the right envelope? And in how many does exactly two?

Choose the correct letters first, then derange the rest. Exactly one correct: $(1 5) D_{4} = 5 \times 9 = 45$ . Exactly two correct: $(2 5) D_{3} = 10 \times 2 = 20$ .

The check worth doing is that the counts over all cases sum to $5!$ : $44 + 45 + 20 + 10 + 0 + 1 = 120$ , where the terms are $D_{5}$ , then exactly one through five correct. The zero is the count for exactly four correct, which is impossible because the last letter would then also be right.

7. Ranking a word

Listing all arrangements of a word's letters in dictionary order and finding the position of one of them is a standard Advanced question, and it is nothing more than a careful place-value count. At each position, count the arrangements that begin with an alphabetically smaller unused letter, then move on.

Illustration 10

Find the rank of MOTHER among the arrangements of its letters in dictionary order.

Sorted, the letters are $E, H, M, O, R, T$ , all distinct.

position	letter	smaller letters still available	contribution
1	M	$E, H$	$2 \times 5! = 240$
2	O	$E, H$	$2 \times 4! = 48$
3	T	$E, H, R$	$3 \times 3! = 18$
4	H	$E$	$1 \times 2! = 2$
5	E	none	$0$
6	R	none	$0$

Adding and including the word itself, the rank is $240 + 48 + 18 + 2 + 1 = 309$ .

If the letters repeat, each contribution uses the multinomial count for the remaining letters instead of a plain factorial — the structure of the calculation does not change.

8. Counting geometric configurations

Points, lines and diagonals give a family of questions whose only difficulty is remembering to subtract the degenerate cases.

Illustration 11

Twelve points lie in a plane, of which five are collinear and no other three are collinear. How many triangles and how many straight lines do they determine?

Triangles: $(3 12) - (3 5) = 220 - 10 = 210$ , since any three of the five collinear points fail to form a triangle.

Lines: $(2 12) - (2 5) + 1 = 66 - 10 + 1 = 57$ , because the ten pairs among the collinear five all give the same single line, which must then be added back once.

Illustration 12

How many diagonals does a convex polygon with $n$ sides have, and for which $n$ does it have exactly $54$ ?

Every pair of vertices gives a segment, and $n$ of those segments are sides, so the count is $(2 n) - n = \frac{n ( n - 3 )}{2}$ . Setting this to $54$ gives $n^{2} - 3 n - 108 = 0$ , so $n = 12$ .

9. Number-formation problems

Digit problems combine the ideas above with two extra constraints: the leading digit cannot be zero, and divisibility conditions usually pin down the last digit or the digit sum.

Illustration 13

How many four-digit numbers with all digits distinct are divisible by $5$ ?

Split by the last digit, since the two cases behave differently.

If the last digit is $0$ , the first three positions take any of the remaining nine digits in order: $9 \times 8 \times 7 = 504$ .

If the last digit is $5$ , the leading digit must avoid both $0$ and $5$ , leaving eight choices, and the middle two positions then have $8$ and $7$ : $8 \times 8 \times 7 = 448$ .

Total $504 + 448 = 952$ . Splitting on the last digit first is essential, because the restriction on the leading digit depends on it.

Illustration 14

Using the digits $1, 2, 3, 4, 5$ without repetition, how many five-digit numbers are divisible by $4$ ?

A number is divisible by $4$ when its last two digits are, so list the two-digit endings available from these digits: $12, 24, 32, 52$ . For each, the remaining three digits fill the first three places in $3! = 6$ ways, giving $4 \times 6 = 24$ .

A related question asks not how many numbers exist but what they add up to. Symmetry does the work: if all four digits $1, 2, 3, 4$ are used without repetition, each digit occupies each of the four places in $3! = 6$ of the arrangements. The units digits therefore contribute $6 \times (1 + 2 + 3 + 4) = 60$ , and the same total appears at every place value, so the sum of all such numbers is

$60 \times 1111 = 66660.$

The argument never lists a single number, and it generalises directly: with $n$ distinct digits the multiplier is $(n - 1)!$ and the repunit has $n$ ones.

Illustration 15

How many six-digit numbers can be formed from the digits of $223355$ that are divisible by $5$ ?

The number must end in $5$ . The remaining five digits are $2, 2, 3, 3, 5$ , arranged in $\frac{5 !}{2 ! 2 !} = 30$ ways. Since no digit is zero, no leading-digit restriction applies, so the answer is $30$ .

Summary

Every count in this chapter is a sum over disjoint, exhaustive cases. The characteristic Advanced error is to satisfy a condition first and then fill the remaining places freely, which over-counts because the finished object cannot remember which part was reserved. Split by the exact number, or count the complement.

Arrangements with repeated objects divide by the factorials of the repeat counts. A "no two together" condition is handled by arranging the other objects first and slotting the restricted ones into the gaps, which makes the separation automatic. Selections from a multiset are counted by shape — how many pairs, how many singletons — because each shape has its own arrangement count.

Distribution questions split four ways depending on whether the objects and the boxes are distinguishable. Identical objects into distinct boxes give stars and bars, $(r - 1 n + r - 1)$ , or $(r - 1 n - 1)$ when no box may be empty; distinct objects into distinct boxes with none empty is the onto count from inclusion and exclusion; and both-identical questions are small enough to list.

Dividing people into groups needs one further reading: labelled groups take the multinomial coefficient as it stands, while unlabelled groups of equal size require dividing by the factorial of how many share a size.

Circular arrangements divide by $n$ for rotations, and by a further $2$ when the object can be turned over. Derangements come from inclusion and exclusion, with $D_{5} = 44$ worth remembering and the recursion worth carrying. Ranking a word is place-value counting with factorials, and geometric and digit problems are ordinary counts with the degenerate cases subtracted.

Key formulas & results

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

Arrangements with repeats

\frac{n !}{n _{1} ! n _{2} ! \dots n _{k} !}

Divide by the shuffles of each block of identical objects. MISSISSIPPI gives $\dfrac{11!}{4!\,4!\,2!}=34650$.

The gap method

arrange the rest, then choose gaps

For "no two together", arrange the unrestricted objects first and place the restricted ones into the gaps, ends included. The separation then holds automatically.

Stars and bars

(r - 1 n + r - 1)

Identical objects into distinct boxes. With every box non-empty it becomes $\binom{n-1}{r-1}$, since one object is placed in each box first.

Selections with repetition

(r n + r - 1)

Choosing $r$ objects from $n$ types with unlimited repetition. It is the same problem as stars and bars, stated as a selection instead of a distribution.

Bounded integer solutions

stars and bars, then inclusion and exclusion

For $x_i\le c$, subtract the solutions with some $x_i\ge c+1$ by substituting $x_i'=x_i-c-1$, then add back the double subtractions.

Onto distributions

\sum_{k} (- 1)^{k} (k r) (r - k)^{n}

Distinct objects into distinct boxes with none empty. For $5$ balls into $3$ boxes it gives $150$; dividing by $3!$ gives $25$ if the boxes are identical.

Division into groups

\frac{n !}{n _{1} ! \dots n _{k} !} \div m!

Divide by $m!$ only when $m$ of the groups have the same size **and** the groups are unlabelled. Nine students into three unlabelled threes give $280$, into three named projects $1680$.

Circular arrangements

(n-1)!

The $n!$ linear orders fall into groups of $n$ rotations. A necklace or garland can also be turned over, halving the count to $\dfrac{(n-1)!}{2}$.

Alternating circular seating

(m - 1)! \times m!

Seat one group circularly to remove the rotation, then fill the gaps with the other. Fixing both groups circularly would remove the rotational freedom twice.

Derangements

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

$D_1=0$, $D_2=1$, $D_3=2$, $D_4=9$, $D_5=44$. The recursion $D_n=(n-1)\left(D_{n-1}+D_{n-2}\right)$ is faster for small $n$.

Exactly $k$ fixed points

(k n) D_{n - k}

Choose which stay fixed, then derange the rest. Summing over all $k$ must return $n!$, which is the check to run.

Rank of a word

1 + \sum_{i} (smaller unused letters) \times (arrangements of the rest)

With repeated letters the second factor is a multinomial rather than a factorial; the structure of the calculation is unchanged.

Diagonals of a polygon

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

All vertex pairs minus the sides. Similarly, lines through points with $m$ collinear are $\binom{n}{2}-\binom{m}{2}+1$.

JEE Advanced test series

Ready to be tested on this?

Permutations and Combinations shows up in every JEE Advanced paper — sit a timed one and find out where you actually stand.

~2Est. Qs

Very highpriority

Full JEE Advanced mocks, real clock
Mathematics-only sectionals & chapter tests
All-India rank + topic-level analysis

Free forever · solver-verified questions

⚠️

Traps JEE Advanced sets — and how to dodge them

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

WATCH OUT

✗ Reserving places to satisfy an "at least" condition, then filling the rest freely

✓ Split by the exact number, or count the complement. Both give

186

for the committee problem, where reserving gives

336

Why it happens: The finished object cannot remember which members were reserved, so a committee with three women is produced once for each pair that could have been the reserved one.

WATCH OUT

✗ Multiplying the total number of selections from a multiset by a single factorial

✓ Count each shape separately, since a selection with a repeated letter has fewer arrangements than one with all letters distinct.

Why it happens: The selection count and the arrangement count are computed in the same breath, and the arrangement factor is silently assumed to be the same for all of them.

WATCH OUT

✗ Using stars and bars when the objects are distinguishable

✓ Ask first whether swapping two objects changes the outcome. If it does, the formula does not apply and the answer is

r^{n}

or an onto count.

Why it happens: The phrase "put

n

things into

r

boxes" is used for all four distribution problems, and only reading the word "identical" separates them.

WATCH OUT

✗ Dividing by the number of groups when the groups are labelled

✓ Divide only when the groups are interchangeable and equal in size. Assigning nine students to three named projects gives

1680

, not

280

Why it happens: The division rule is memorised alongside the multinomial coefficient without its condition, so it gets applied whenever equal group sizes appear.

WATCH OUT

✗ Treating a garland like a dinner table

✓ Halve the circular count whenever the arrangement can be turned over, giving

\frac{( n - 1 )!}{2}

for necklaces and garlands.

Why it happens: Both are described as circular arrangements, and the physical difference — that a garland has no fixed face — is not part of the wording candidates attend to.

WATCH OUT

✗ Fixing both groups when seating two sets alternately

✓ Remove the rotational freedom once, by seating one group circularly, and arrange the other linearly into the gaps.

Why it happens: The rule "circular arrangements use

(n - 1)!

" is applied to each group in turn, dividing by the rotation twice and undercounting by a factor of

n

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 Permutations and Combinations?

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 ~8 marks in JEE Advanced exams

5-minute revision

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

•Every valid count is a sum over disjoint, exhaustive cases; reserving places to meet a condition and filling freely always over-counts.
•The complement is often one term where the direct count is four; check whether it is shorter before starting.
•Arrangements with repeats divide by the factorial of each repeat count.
•For "no two together", arrange the others first and slot the restricted objects into the gaps, ends included.
•Selections from a multiset are counted by shape, because each shape carries its own arrangement factor.
•Stars and bars gives $(r - 1 n + r - 1)$ ; with no box empty it is $(r - 1 n - 1)$ .
•Upper bounds on the variables are handled by subtracting shifted solutions and adding back the double subtractions.
•Distributions split four ways; only identical-into-distinct has a clean formula, and both-identical is listed by hand.
•Divide a grouping count by $m!$ only when $m$ groups are the same size and the groups are unlabelled.
•Circular arrangements are $(n - 1)!$ , and halve again for anything that can be turned over.
• $D_{5} = 44$ ; exactly $k$ correct is $(k n) D_{n - k}$ , and the cases must sum to $n!$ .
•Ranking a word is place-value counting: at each position, count the arrangements starting with a smaller unused letter.

JEE Advanced question blueprint

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

Typical weightage: ~2 questions (roughly 6-8 marks) across the two papers combined, out of the ~120 marks of Mathematics

Question style	Marks each	Typical count	What it tests
Arrangements, selections and restricted positions	3	1	Arrangements with repeated objects, separation conditions by the gap method, selections from a multiset, and restricted committee formation
Distributions, groupings and integer solutions	4	1	The four distribution types, stars and bars with lower and upper bounds, onto distributions, and division into labelled or unlabelled groups
Derangements, ranking and applied counting	3	1	Derangements and exact fixed-point counts, the rank of a word, counting geometric configurations, and number-formation with divisibility conditions

Exam-hall strategy

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

Before computing anything, write down the cases you intend to add. If they overlap, the answer is already wrong, and the overlap is easier to see at that stage than afterwards.
For any "at least" condition, evaluate the complement count as well. It costs a few seconds and catches the single most common error in the chapter.
Ask explicitly whether each set of objects and each set of positions is distinguishable. Most distribution questions are decided by that one reading, not by a formula.
In circular problems, remove the rotational freedom exactly once, and check whether the object can be turned over before halving.
Where a small case can be listed, list it. A count of five or six checked by hand validates the method you are about to apply to the large case.

Beyond the exam

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

Cryptographic key strength is a permutation count: the pr…

Cryptographic key strength is a permutation count: the practical security of a password comes from the size of the space of possible strings, not from any property of the particular string chosen.

Clinical trial designers use combinatorial balancing to a…

Clinical trial designers use combinatorial balancing to assign treatments across groups so that every pairing of factors appears equally often, which is a distribution problem with labelled boxes.

Network routing algorithms bound their own running time b…

Network routing algorithms bound their own running time by counting paths through a graph, and the derangement idea appears whenever a schedule must avoid returning any item to its original position.

Where else this topic is tested

Prepare once, score in every exam that asks it.

JEE Advanced

JEE Main

BITSAT

Mathematics Olympiad (regional level)

CAT (Quantitative Ability)

📝 My notes

Saved on this device — sign in to sync · how this works

0/600

Questions aspirants ask

Pulled from the Q&A community and mentor sessions.

Because the finished committee carries no record of which members were reserved. A committee with three women is generated once for each of the three pairs that could have been chosen at the reservation step, so it is counted three times. The method would be valid if the reserved seats were distinguishable from the others — for instance if two specific posts had to be held by women — and that is exactly the variant Advanced sets to test whether the distinction is understood.

Count the number of cases each route needs. "At least one" almost always has a single-term complement, and "at least two out of four" has three direct cases against two complementary ones. When the two routes are close in length, do both: they are quick, and agreement is a genuine check on a chapter where a plausible wrong answer is the usual failure mode.

Only when the boxes are genuinely interchangeable. Five distinct balls into three distinct boxes with none empty is $150$ ; if the boxes are identical, the same partition has been counted once for each of the $3!$ labellings, so the answer is $25$ . The division is valid here because every distribution uses all three boxes; if empty boxes were allowed, some arrangements would have fewer distinct blocks and the simple division would fail.

No usable one. That count is the number of partitions of an integer into at most a given number of parts, and it has no closed form. This is deliberate: questions of that type are always set with small numbers, so listing the partitions is the intended method. Five identical balls into three identical boxes gives five arrangements, found by writing them all down.

A large part of it, in effect. Most probability questions on the paper are counting questions with a division at the end, so an error in the numerator or the denominator here costs marks in two places. It is worth treating any counting slip found in this chapter as a probability slip as well, and re-checking the equally-likely assumption that turns a count into a probability.

More Mathematics chapters

Sets, Relations and FunctionsCounting relations on a finite set, equivalence classes as partitions, ranges found by inv…

Complex Numbers and Quadratic EquationsMultiplication as rotation, roots of unity as a polygon, loci that are arcs rather than ci…

Matrices and DeterminantsDeterminants as area factors, the factor theorem applied to a determinant, adjoint identit…

Binomial TheoremThe general term as the only tool, greatest term against greatest coefficient, coefficient…

Sources and How This Chapter Was CheckedSyllabus scope, what was derived rather than quoted, and how every answer here was checked.

Scope follows the JEE Advanced syllabus for 2026 (Mathematics, Permutations and Combinations): the fundamental principle of counting, permutations and combinations of distinct and non-distinct objects, arrangements in a line and in a circle, and their applications.

The treatment concentrates on what Advanced adds to Main. Main asks for a single permutation or combination count; Advanced asks for a count under two interacting restrictions, for a distribution whose answer depends on which objects are distinguishable, for a derangement condition, and for the rank of a word.

Results were derived rather than quoted. The multinomial coefficient came from dividing by the shuffles of identical objects, the stars-and-bars formula from encoding a distribution as a sequence of symbols, the derangement formula from inclusion and exclusion over fixed positions, and the diagonal count from subtracting the sides from all vertex pairs.

Every illustration was checked a second way. The opening committee count was obtained twice, by cases and by complement; the bounded solutions of $x + y + z = 10$ were confirmed by listing all six; the derangement counts were verified by checking that the exactly- $k$ -correct totals sum to $5!$ ; and the onto distribution reused the value computed independently in the chapter on functions.

The illustrations are teaching problems written for this chapter, not previous-year questions, and are not labelled as such.

Take the next step

JEE Advanced Mathematics — find a tutor

Pair self-study with a tutor, a live course or a coaching centre.

1-on-1 help

🧑‍🏫

Find a verified tutor

School, college and competitive — online or at home.

Find tutors

Live cohorts

Join a live course

Test series and recorded lessons from top creators.

Browse courses

Classroom

Browse coaching centres

Verified institutes near you for every exam.

Find centres

Permutations and Combinations

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

Before you start — revise these

Permutations and Combinations

1. Arrangements when some objects are alike

Illustration 1

Illustration 2

2. Selecting from a multiset

Illustration 3

3. Distributions: four different questions

Illustration 4

Illustration 5

Illustration 6

4. Division into groups, and when to divide by the number of groups

Illustration 7

5. Circular arrangements

Illustration 8

6. Derangements

Illustration 9

7. Ranking a word

Illustration 10

8. Counting geometric configurations

Illustration 11

Illustration 12

9. Number-formation problems

Illustration 13

Illustration 14

Illustration 15

Summary

Key formulas & results

Traps JEE Advanced sets — and how to dodge them

Exam-pattern practice

Are you exam-ready for Permutations and Combinations?

5-minute revision

JEE Advanced question blueprint

Exam-hall strategy

Beyond the exam

Where else this topic is tested

📝 My notes

Questions aspirants ask

Why exactly does the "reserve two women first" method fail?

How do I decide whether to count directly or by complement?

When does a distribution problem need division by the number of boxes?

Is there a formula for putting identical objects into identical boxes?

How much of this chapter is really about probability?

More Mathematics chapters

JEE Advanced Mathematics — find a tutor

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities