JEE Maths — Permutations and Combinations Complete Chapter Guide

Chapter Overview & Weightage

Permutations and Combinations (P&C) is one of those chapters where a single concept — “is order important?” — decides everything. Get that distinction sharp, and the chapter becomes very manageable.

P&C consistently contributes 1–2 questions in JEE Main, translating to 4–8 marks per paper. In JEE Advanced, it often appears as part of a larger combinatorics problem worth 3–4 marks. Overall weightage sits at 4–5% — not the heaviest chapter, but highly scorable because the question types are predictable.

Year	JEE Main (Questions)	JEE Advanced (Questions)	Topics Covered
2024	2	1	Selections with restrictions, circular arrangements
2023	1	1	Derangements, word formation
2022	2	2	Multinomial, distribution problems
2021	1	1	Divisors, committee selection
2020	2	1	Rank of a word, restricted arrangements

The pattern is clear: JEE Main loves word formation and selection with conditions. JEE Advanced prefers derangements and distribution problems that require multi-step reasoning.

Key Concepts You Must Know

Prioritised by how often they appear in PYQs:

Fundamental Counting Principle — the foundation. If task A can be done in $m$ ways and task B in $n$ ways independently, together they can be done in $m \times n$ ways.
Permutations $^nP_r$ — arrangements where order matters. Picking and placing $r$ objects from $n$ distinct objects.
Combinations $^nC_r$ — selections where order does NOT matter. Choosing $r$ from $n$ .
Permutations with repetition — arranging $n$ objects where some are identical. Divide by factorials of repeating groups.
Circular arrangements — $n$ distinct objects in a circle: $(n-1)!$ ways. Necklace/bracelet: $(n-1)!/2$ .
Restricted arrangements — “always together” (treat as a block) vs “never together” (subtract bad cases).
Selections with conditions — at least one, at most one, exactly $k$ from a group.
Derangements — arrangements where no object occupies its original position. Formula is critical for JEE Advanced.
Distribution problems — distributing $n$ identical/distinct objects into $r$ identical/distinct boxes.
Rank of a word in dictionary — find the position of a given arrangement among all alphabetical arrangements.

Important Formulas

^nP_r = \frac{n!}{(n-r)!}

^nC_r = \frac{n!}{r!(n-r)!}

Identical objects arrangement (multinomial):

\frac{n!}{p! \cdot q! \cdot r! \cdots}

where $p, q, r, \ldots$ are counts of each identical group.

When to use $^nP_r$ vs $^nC_r$ : Ask yourself — does swapping two selected items give a different outcome? If yes (like seating positions), use $P$ . If no (like a committee), use $C$ .

Distinct objects in a circle: $(n-1)!$

Necklace/bracelet (can be flipped): $\dfrac{(n-1)!}{2}$

Why $(n-1)!$ and not $n!$ ? In a circle, one position is fixed to remove rotational equivalence. You’re essentially arranging the remaining $n-1$ people relative to a fixed reference.

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

For small values (memorise these):

$n$	$D_n$
1	0
2	1
3	2
4	9
5	44

Recurrence: $D_n = (n-1)(D_{n-1} + D_{n-2})$

$n$ distinct objects into $r$ distinct boxes (empty boxes allowed):

r^n

$n$ identical objects into $r$ distinct boxes (empty boxes allowed):

^{n+r-1}C_{r-1}

$n$ identical objects into $r$ distinct boxes (no empty box):

^{n-1}C_{r-1}

The “stars and bars” formula $^{n+r-1}C_{r-1}$ is the most frequently tested distribution result in JEE Main. It models situations like “how many ways to give 10 identical chocolates to 4 students” — a recurring problem structure.

If $N = p_1^{a_1} \cdot p_2^{a_2} \cdots p_k^{a_k}$ , the number of divisors is:

(a_1 + 1)(a_2 + 1) \cdots (a_k + 1)

Solved Previous Year Questions

PYQ 1 — JEE Main 2024 Shift 1

Q. The number of words (with or without meaning) that can be formed using all the letters of the word PERMUTATION such that the vowels always occur together is:

Solution:

The word PERMUTATION has 11 letters: P, E, R, M, U, T, A, T, I, O, N.

Vowels: E, U, A, I, O → 5 vowels (all distinct). Consonants: P, R, M, T, T, N → 6 consonants, with T repeated twice.

Step 1: Treat the 5 vowels as a single block. Now we arrange: [VOWELS], P, R, M, T, T, N → 7 units total.

Step 2: Arrange these 7 units. The T repeats, so:

\frac{7!}{2!} = \frac{5040}{2} = 2520

Step 3: Arrange the 5 vowels within the block. All distinct:

5! = 120

Total $= 2520 \times 120 = \mathbf{302400}$

Many students forget to divide by $2!$ for the repeated T in consonants. The vowel block is treated as one unit — the T’s are among the consonants, not inside the block. Always list out identical letters carefully before counting.

PYQ 2 — JEE Main 2023 January Session

Q. The number of 4-letter words (with or without meaning) that can be formed from the letters of the word IMPOSSIBLE such that no letter is repeated, is:

Solution:

Letters in IMPOSSIBLE: I, M, P, O, S, S, I, B, L, E

Distinct letters: I, M, P, O, S, B, L, E → 8 distinct letters (note: I appears twice, S appears twice, but for “no repetition” we treat each as one available type).

We choose 4 from these 8 distinct types and arrange:

^8P_4 = 8 \times 7 \times 6 \times 5 = \mathbf{1680}

When a problem says “no letter repeated,” work only with the distinct letters available, ignoring how many times each appears in the original word. The repetitions in the source word only matter when repetition IS allowed.

PYQ 3 — JEE Advanced 2022

Q. Five persons A, B, C, D, E are seated in a circular arrangement. If A and B must not sit adjacent to each other, find the number of arrangements.

Solution:

Step 1: Total circular arrangements of 5 persons: $(5-1)! = 24$

Step 2: Count arrangements where A and B are adjacent. Treat A and B as a single block → 4 units in a circle:

$(4-1)! = 6$ circular arrangements of the block.

Within the block, A and B can swap: $\times 2 = 12$ arrangements.

Step 3: Subtract:

24 - 12 = \mathbf{12}

Difficulty Distribution

For JEE Main and Advanced:

Difficulty	Weightage	Typical Topics
Easy (direct formula)	~40%	Basic $^nP_r$ , $^nC_r$ , word formation without restrictions
Medium (one condition)	~40%	Restricted arrangements, selection with conditions, rank of word
Hard (multi-step / derangements)	~20%	Derangements, distribution with constraints, advanced circular

JEE Main typically asks 1 easy + 1 medium question from this chapter. JEE Advanced, when it appears, is always a hard problem — usually derangements or an unusual distribution setup. Do not spend more than 15 minutes on a hard P&C problem in Advanced; the chapter’s weightage doesn’t justify it.

Expert Strategy

Phase 1: Build the Core (Week 1)

Start with fundamental counting and work through $^nP_r$ , $^nC_r$ problems until “is order important?” becomes a reflex. Do at least 20 straightforward problems before moving to conditions.

Phase 2: Master Conditions (Week 2)

The bulk of JEE questions involve at least one condition: always together, never together, exactly $k$ women, at most 2 from a group. Practice each condition type in isolation, then mix them.

For “never together” problems: subtract the “always together” count from total — don’t try to count directly. Direct counting leads to over-counting or missed cases almost every time.

Phase 3: PYQs + Special Topics (Week 3)

Spend two sessions on rank of a word problems — the technique is fixed and predictable. One session on derangements (memorise $D_3 = 2$ , $D_4 = 9$ , the recurrence). One session on distribution using stars and bars.

During the Exam

Read the question twice and identify:

Is order relevant? → $P$ or $C$ ?
Are there identical objects? → Divide by repeated factorials.
Are there conditions? → Break into cases or use complement.
Is it a circular arrangement? → Fix one element.

This 30-second checklist prevents the most common errors.

Common Traps

Trap 1 — Circular vs Linear: A question says “arranged around a table” but students apply $n!$ instead of $(n-1)!$ . If the problem says “in a row” or “on a shelf,” use $n!$ . If it says “around a table,” “in a circle,” or “in a ring,” use $(n-1)!$ .

Trap 2 — The “at least one” overcounting trap: For “at least one from group A,” students often try to add cases (exactly 1 + exactly 2 + …). The clean method is: Total selections − selections with no member from A. Saves time and reduces errors.

Trap 3 — Identical vs Distinct boxes: “Distribute 10 identical balls into 4 identical boxes” is different from “4 distinct boxes.” JEE Main almost always specifies distinct boxes (stars and bars applies). Identical boxes require Partition Theory — rarely tested in Main, occasionally in Advanced.

Trap 4 — The necklace flip: A necklace can be flipped (no front/back). So arrangements that are mirror images of each other count as the same. Always divide $(n-1)!$ by 2 for necklace problems. A garland is also a necklace. A seating arrangement around a table is NOT — don’t divide by 2 there.

Trap 5 — Rank of a word starting letter: When finding the rank of a word, students sometimes forget to account for letters smaller than the first letter. List all distinct letters alphabetically, count how many come before each position, multiply by the arrangements possible with remaining letters. One missed letter shifts the rank by hundreds.

A specific pattern from JEE Main 2022 and 2023: problems asking for the number of ways to form committees with exactly $k$ women where the group has specified numbers of men and women. The answer is always $^{w}C_k \times ^{m}C_{n-k}$ where $w$ = women available, $m$ = men available, $n$ = committee size. Recognise this template immediately — it’s a 90-second solve.