JEE Main Maths Permutations & Combinations 2027: Counting Principles, Arrangements & Selections

Permutations and Combinations has almost no formulas to learn and yet it feels hard, because every question is a fresh puzzle. The trick is a fixed routine: decide whether order matters, list the restrictions, then count in stages. Once the routine is a habit, the puzzles stop looking random.

Ask one question first: does the order matter? The whole chapter branches from that answer.

---

Chapter at a Glance

Priority reflects past-paper trends, not an official NTA weightage.

What the NTA Syllabus Covers

• The fundamental principle of counting
• Permutations and combinations, meaning of $P(n, r)$ and $C(n, r)$
• Simple applications

---

Master These Topics

1. The Fundamental Principle of Counting

If one task can be done in $m$ ways and a second independent task in $n$ ways, the two together can be done in $m \times n$ ways (multiplication principle). If the tasks are alternatives, the total is $m + n$ (addition principle).

2. Permutations and Combinations

• Arrangements of $r$ objects out of $n$ distinct objects: $P(n, r) = \dfrac{n!}{(n - r)!}$
• Selections of $r$ objects out of $n$: $C(n, r) = \dbinom{n}{r} = \dfrac{n!}{r!\,(n - r)!}$
• Relation: $P(n, r) = r!\, C(n, r)$, and $C(n, r) = C(n, n - r)$

Worked example (selection with a condition): A committee of 3 is chosen from 5 men and 4 women with at least one woman. Total selections are $C(9, 3) = 84$, and those with no woman are $C(5, 3) = 10$. So the answer is $84 - 10 = 74$.

3. Arrangements with Repeated Objects and in a Circle

• $n$ objects with $p$ alike of one kind, $q$ alike of another: $\dfrac{n!}{p!\,q!}$
• Arrangement in a circle of $n$ distinct objects: $(n - 1)!$

Worked example (repeated letters): MISSISSIPPI has 11 letters with I appearing 4 times, S 4 times and P 2 times, so the number of arrangements is $\dfrac{11!}{4!\,4!\,2!} = 34650$.

Worked example (circle): Six people around a round table can sit in $5! = 120$ ways. If two particular people must sit together, treat them as one block: $2 \times 4! = 48$ ways.

Trap: In circular arrangements, rotations count as the same seating, which is why we fix one object and arrange the rest.

4. Restrictions: The Gap Method

To keep certain objects apart, first arrange the others and then place the restricted objects in the gaps.

Worked example: Arrange 5 boys and 3 girls in a row so that no two girls sit together. Arrange the boys in $5!$ ways, which creates 6 gaps. Choose 3 gaps for the girls in $C(6, 3)$ ways and arrange them in $3!$ ways:

$$5! \times C(6, 3) \times 3! = 120 \times 20 \times 6 = 14400$$

5. Geometry Counting

• Diagonals of an $n$-sided polygon: $\dfrac{n(n - 3)}{2}$. For $n = 10$ this gives $35$.
• Triangles from 8 points of which 3 are collinear: $C(8, 3) - C(3, 3) = 56 - 1 = 55$.

---

Common Traps to Avoid

• Using a permutation when the problem asks only for a selection, or the reverse.
• Forgetting to divide by $p!$ when objects are identical.
• Counting rotations of a circular arrangement as different.
• Adding when you should multiply. Independent stages multiply, alternatives add.

60-Second Revision Sheet

• $P(n, r) = \dfrac{n!}{(n-r)!}$, $C(n, r) = \dfrac{n!}{r!(n-r)!}$
• Repeated objects: $\dfrac{n!}{p!\,q!\cdots}$; circle: $(n-1)!$
• Diagonals of an $n$-gon: $\dfrac{n(n-3)}{2}$
• Complement trick: total minus the unwanted cases

Your Study Plan

1. Day 1: multiplication and addition principles with simple word problems.
2. Day 2: permutations with and without repetition.
3. Day 3: combinations, committee problems and the complement trick.
4. Day 4: circular arrangements, gap method and a timed mixed set.

Practice Permutations & Combinations Questions Free → (opens in a new tab)

---

Continue Your Maths Journey

• Previous chapter: Matrices & Determinants
• Next chapter: Binomial Theorem
• All 14 JEE Main Maths chapters
• Complete JEE Main Syllabus 2027 guide

---

Frequently Asked Questions