Permutation and Combination Formulas for JEE Main and Advanced

MathsClass 11, Permutations and CombinationsFor JEE Main and Advanced

Permutation and combination formulas for JEE Main and Advanced on one page: the counting principles, $^{n} P_{r}$ and $^{n} C_{r}$ with their properties, arrangements with repeats, blocks, gaps and circles, selections with points and lines, integer solutions, distributions, derangements and counting functions, each with the condition under which it holds. The table shows which formulas decided the most questions in recent JEE papers, and section 6 solves ten of them step by step. Permutations and combinations is a Class 11 chapter, and its counting comes back in probability and the binomial theorem.

Which formulas JEE uses most

We read all 55 questions on permutation and combination in our 48 past papers (JEE Main 2025-2026, JEE Advanced 2022-2026) and noted the formulas each solution needs; one question can need several. The formulas used most:

Formula	Questions
$^{n} C_{r} = \frac{n !}{r ! ( n - r ) !}$ with $^{n} C_{r} =^{n} C_{n - r}$ and Pascal's rule $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$	2545%
Arrangements with repeated objects: $\frac{n !}{p ! q ! s !}$	1120%
Total minus unwanted, inclusion-exclusion, the block method for "together"	1120%
Fundamental principle of counting: multiply for stages, add for cases	713%
$^{n} P_{r} = \frac{n !}{( n - r ) !}$	59%
Points and lines: triangles $^{n} C_{3} -^{k} C_{3}$ when $k$ are collinear, diagonals $^{n} C_{2} - n$	59%
Functions from an $n$ -set to an $m$ -set: $m^{n}$ in all, $^{m} P_{n}$ one-one, $^{m} C_{n}$ strictly increasing	59%
Rank of a word in dictionary order	35%

Show the other 7

Formula	Questions
$x_{1} + x_{2} + \dots + x_{r} = n$ with every $x_{i} \geq 0$ : $^{n + r - 1} C_{r - 1}$ solutions	35%
Power of a prime $p$ in $n!$ : $[\frac{n}{p}] + [\frac{n}{p ^{2}}] + [\frac{n}{p ^{3}}] + \dots$	24%
Repetition allowed: $n^{r}$	24%
Gap method for "no two together"; no two consecutive from $1, 2, \dots, n$ : $^{n - r + 1} C_{r}$	24%
Number of subsets of an $n$ -set: $2^{n}$	12%
Distinct objects into distinct boxes with none empty (inclusion-exclusion)	12%
Dividing into groups of sizes $p, q, s$ : $\frac{n !}{p ! q ! s !}$	12%

1. Counting principles, $^{n} P_{r}$ and $^{n} C_{r}$

Fundamental principle of counting

stages: m \times n cases: m + n

If one step can be done in $m$ ways and the next step in $n$ ways, both steps together can be done in $mn$ ways. If the job is done one way or another way (separate cases that cannot happen together), add the counts.

Expression	Equals	Note
$^{n} P_{r}$	$\frac{n !}{( n - r ) !}$	Arrangements of $r$ out of $n$ different objects: the order counts
$^{n} C_{r}$	$\frac{n !}{r ! ( n - r ) !} = \frac{^{n} P _{r}}{r !}$	Selections: the order does not count
$^{n} C_{r}$	$^{n} C_{n - r}$	So $^{n} C_{a} =^{n} C_{b}$ gives $a = b$ or $a + b = n$
$^{n} C_{r} +^{n} C_{r - 1}$	$^{n + 1} C_{r}$	Pascal's rule
$^{n} C_{r}$	$\frac{n}{r}^{n - 1} C_{r - 1}$	Takes a factor out of a sum
$\frac{^{n} C _{r}}{^{n} C _{r - 1}}$	$\frac{n - r + 1}{r}$	So $^{n} C_{r}$ is largest in the middle: at $r = \frac{n}{2}$ for even $n$ , at $r = \frac{n \pm 1}{2}$ for odd $n$
$^{r} C_{r} +^{r + 1} C_{r} + \dots +^{n} C_{r}$	$^{n + 1} C_{r + 1}$	Hockey-stick sum
$^{n} C_{0} +^{n} C_{1} + \dots +^{n} C_{n}$	$2^{n}$	The number of subsets of an $n$ -set

2. Arrangements

Arrangement	Number	Note
In a row	$n!$	$n$ different objects
Repetition allowed	$n^{r}$	$r$ places, each filled from the same $n$ different objects
Repeated objects	$\frac{n !}{p ! q ! s !}$	$n$ objects in a row: $p$ alike of one kind, $q$ alike of another, $s$ alike of a third
Fixed order	$\frac{n !}{k !}$	$k$ of the $n$ objects must come in a given order (the vowels in alphabetical order, say)
Together: the block method	$(n - k + 1)! k!$	$k$ given objects side by side: glue them into one block, arrange the $n - k + 1$ units, then the $k$ inside the block
No two together: the gap method	$(n - k)! \cdot^{n - k + 1} P_{k}$	Arrange the other $n - k$ first, then put the $k$ objects in different gaps out of the $n - k + 1$ (ends included). Needs $k \leq n - k + 1$ .
Around a circle	$(n - 1)!$	Turning the whole circle gives the same arrangement
Necklace or garland	$\frac{( n - 1 ) !}{2}$	Turning it over also gives the same one ( $n \geq 3$ )
$r$ of $n$ around a circle	$\frac{^{n} P _{r}}{r}$	Rotations count as the same; as a necklace it is $\frac{^{n} P _{r}}{2 r}$ ( $r \geq 3$ )

Rank of a word in dictionary order

rank = (words before it) + 1

Write the letters in alphabetical order. Go along the word from the left: at each place, count the words that keep the letters already fixed and put a smaller letter here (arrange the letters still left, dividing by $p!$ for a letter left $p$ times). Add the counts.

3. Selections, points and lines

Selection	Number	Note
Subsets	$2^{n}$	Each of $n$ different objects is in or out; at least one object: $2^{n} - 1$
From alike objects	$(p + 1) (q + 1) (s + 1) - 1$	$p$ alike, $q$ alike and $s$ alike objects of three kinds: take any number of each, at least one object in all
Divisors of $N$	$(a + 1) (b + 1) (c + 1)$	$N = p_{1}^{a} p_{2}^{b} p_{3}^{c}$ in primes; $1$ and $N$ are counted. Their sum: $(1 + p_{1} + \dots + p_{1}^{a}) \dots (1 + p_{3} + \dots + p_{3}^{c})$
No two consecutive	$^{n - r + 1} C_{r}$	$r$ numbers from $1, 2, \dots, n$ with no two consecutive
Lines through $n$ points	$^{n} C_{2} -^{k} C_{2} + 1$	$k \geq 2$ of the points on one line and no other three on a line
Triangles from $n$ points	$^{n} C_{3} -^{k} C_{3}$	The same set-up; with no three on a line it is $^{n} C_{3}$
Diagonals of an $n$ -gon	$^{n} C_{2} - n = \frac{n ( n - 3 )}{2}$
Triangles on the corners of an $n$ -gon	$^{n} C_{3}$	With no side along a side of the polygon: $\frac{n ( n - 4 ) ( n - 5 )}{6}$ ( $n \geq 4$ )
Rectangles in a grid	$^{m} C_{2} \cdot^{n} C_{2}$	$m$ horizontal and $n$ vertical lines. An $n \times n$ board has $1^{2} + 2^{2} + \dots + n^{2}$ squares.

4. Distributions, integer solutions and functions

Distribution	Number	Note
Identical objects, $x_{i} \geq 0$	$^{n + r - 1} C_{r - 1}$	Solutions of $x_{1} + x_{2} + \dots + x_{r} = n$ , which is $n$ identical objects into $r$ different boxes with empty boxes allowed
Identical objects, $x_{i} \geq 1$	$^{n - 1} C_{r - 1}$	No box empty
Upper limit $x_{i} \leq m$	$\sum_{j \geq 0} (- 1)^{j} \cdot^{r} C_{j}^{n - j (m + 1) + r - 1} C_{r - 1}$	Inclusion-exclusion: take away the solutions with some $x_{i} \geq m + 1$ ; stop when $n - j (m + 1) < 0$
Different objects, any box	$r^{n}$	$n$ different objects into $r$ different boxes
Different objects, no box empty	$\sum_{j = 0}^{r} (- 1)^{j} \cdot^{r} C_{j} (r - j)^{n}$	For $r = 3$ boxes: $3^{n} - 3 \cdot 2^{n} + 3$
Different objects, identical boxes, none empty	$\frac{1}{r !} \sum_{j = 0}^{r} (- 1)^{j} \cdot^{r} C_{j} (r - j)^{n}$	The count above divided by $r!$ : $4$ different books into $3$ identical bags, none empty, in $\frac{36}{3 !} = 6$ ways. Identical objects into identical boxes: list the ways to split $n$ by hand.
Groups of sizes $p$ , $q$ , $s$	$\frac{n !}{p ! q ! s !}$	$p + q + s = n$ , sizes all different, or the groups named (teams X, Y and Z). If $t$ unnamed groups have the same size, divide by $t!$ : sizes $2, 2, 3$ give $\frac{7 !}{2 ! 2 ! 3 ! 2 !} = 105$ .
Equal groups	$\frac{( mk ) !}{( k !) ^{m} m !}$	$mk$ objects into $m$ unnamed groups of $k$ ; given out to $m$ people it is $\frac{( mk ) !}{( k !) ^{m}}$
Derangements	$D_{n} = n! (1 - \frac{1}{1 !} + \frac{1}{2 !} - \dots + \frac{( - 1 ) ^{n}}{n !})$	No object in its own place: $D_{2} = 1$ , $D_{3} = 2$ , $D_{4} = 9$ , $D_{5} = 44$

For functions from a set $A$ with $n$ elements to a set $B$ with $m$ elements:

Functions	Number	Note
All functions	$m^{n}$
One-one	$^{m} P_{n}$	Needs $n \leq m$
Onto	$\sum_{j = 0}^{m} (- 1)^{j} \cdot^{m} C_{j} (m - j)^{n}$	Needs $n \geq m$
Strictly increasing	$^{m} C_{n}$	Sets of numbers: choose the $n$ values and the order is forced
Non-decreasing	$^{m + n - 1} C_{n}$	Values may repeat: a selection with repetition

Power of a prime

p

n!

[\frac{n}{p}] + [\frac{n}{p ^{2}}] + [\frac{n}{p ^{3}}] + \dots

$[x]$ is the greatest integer function; stop once $p^{k} > n$ . The number of zeros at the end of $n!$ is the power of $5$ in it.

5. Exam traps

Watch out

Order matters (arrangements, numbers, queues): $^{n} P_{r}$ . Order does not matter (committees, teams, selections): $^{n} C_{r}$ .
Identical or different objects, identical or different boxes: four different counts. Decide this first.
"Not all together" is total minus all together; "no two together" is the gap method. They are different questions.
A number cannot start with $0$ : take away the arrangements that do.
Around a circle with unmarked seats it is $(n - 1)!$ , not $n!$ (numbered seats: $n!$ ). Divide by $2$ only when turning over gives the same arrangement (necklace, garland).
Equal groups that are not named: divide by $m!$ . Named groups or groups given to people: do not.
In dictionary rank, sort the letters first and divide by $p!$ for repeated letters still left.
With $k$ collinear points: $^{k} C_{3}$ triangles are lost, and $^{k} C_{2}$ lines become $1$ line.

6. Past-paper questions, solved

Ten questions from our past papers, chosen to cover the formulas used most. The question and the answer key are as set in the paper; the working is ours.

PYQ 1: Total minus unwanted

JEE Main 2026 Apr 4 Shift 1, Mathematics Q8

The number of ways, of forming a queue of $4$ boys and $3$ girls such that all the girls are not together, is:

(A) $5040$
(B) $3050$
(C) $3410$
(D) $4320$

Solution:

All queues of the $7$ people: $7! = 5040$ .
Queues with all $3$ girls together: glue them into one block, arrange the $5$ units in $5!$ ways and the girls inside the block in $3!$ ways: $5! \cdot 3! = 720$ .
Girls not all together: $5040 - 720 = 4320$ . (In the question, "all the girls are not together" means not all three together.)

Answer key (D) $4320$

PYQ 2: Block method and gap method

JEE Main 2025 Jan 23 Shift 2, Mathematics Q21

The number of ways, $5$ boys and $4$ girls can sit in a row so that either all the boys sit together or no two boys sit together, is ________.

Solution:

All $5$ boys together: the block of boys and the $4$ girls make $5$ units, arranged in $5!$ ways; the boys inside the block in $5!$ ways. That gives $5! \cdot 5! = 14400$ .
No two boys together: seat the $4$ girls in $4!$ ways. They leave $5$ gaps (the two ends included), one boy in each: $5!$ ways. That gives $4! \cdot 5! = 2880$ .
The two cases cannot happen together, so add them: $14400 + 2880 = 17280$ .

Answer key 17280

PYQ 3: Dictionary rank

JEE Main 2026 Jan 24 Shift 2, Mathematics Q9

The letters of the word "UDAYPUR" are written in all possible ways with or without meaning and these words are arranged as in a dictionary. The rank of the word "UDAYPUR" is :

(A) 1580
(B) 1578
(C) 1579
(D) 1581

Solution:

Letters in alphabetical order: A, D, P, R, U, U, Y (U twice).
Words starting with A, D, P or R: each leaves $6$ letters with U twice, so $\frac{6 !}{2 !} = 360$ words each, $1440$ in all.
Words starting with U: the letters left are A, D, P, R, U, Y, all different. A second letter smaller than D is only A: $5! = 120$ words. Total so far $1560$ .
After UD the next letter A is the smallest left: no words. After UDA the letters left are P, R, U, Y, and P, R and U are smaller than Y: $3 \cdot 3! = 18$ words. Total $1578$ .
After UDAY the next letter P is the smallest left. After UDAYP the letters left are R and U, and R is smaller than U: $1$ word (UDAYPRU). Total $1579$ .
So UDAYPUR is word number $1579 + 1 = 1580$ .

Answer key (A) 1580

PYQ 4: Arrangements with repeated digits

JEE Main 2026 Apr 2 Shift 1, Mathematics Q6

The number of seven-digit numbers, that can be formed by using the digits $1, 2, 3, 5$ and $7$ such that each digit is used at least once, is :

(A) $15400$
(B) $17800$
(C) $16800$
(D) $29400$

Solution:

Seven places, five digits, each digit at least once: the digit counts are $3, 1, 1, 1, 1$ (one digit three times) or $2, 2, 1, 1, 1$ (two digits twice).
One digit three times: choose it in $5$ ways and arrange the seven digits in $\frac{7 !}{3 !} = 840$ ways: $4200$ .
Two digits twice: choose them in $^{5} C_{2} = 10$ ways and arrange in $\frac{7 !}{2 ! 2 !} = 1260$ ways: $12600$ .
Total: $4200 + 12600 = 16800$ . The onto formula agrees: $5^{7} - 5 \cdot 4^{7} + 10 \cdot 3^{7} - 10 \cdot 2^{7} + 5 = 16800$ .

Answer key (C) $16800$

PYQ 5: Triangles from points with some collinear

JEE Main 2025 Apr 8 Shift 2, Mathematics Q13

There are 12 points in a plane, no three of which are in the same straight line, except 5 points which are collinear. Then the total number of triangles that can be formed with the vertices at any three of these 12 points is

(A) $230$
(B) $220$
(C) $200$
(D) $210$

Solution:

Any $3$ of the $12$ points: $^{12} C_{3} = 220$ choices.
Three of the $5$ collinear points give no triangle: $^{5} C_{3} = 10$ choices.
Triangles: $220 - 10 = 210$ .

Answer key (D) $210$

PYQ 6: Integer solutions (stars and bars)

JEE Main 2026 Jan 23 Shift 2, Mathematics Q15

The number of ways, in which 16 oranges can be distributed to four children such that each child gets at least one orange, is

(A) 429
(B) 384
(C) 403
(D) 455

Solution:

Take the oranges as identical (the usual reading; the key $455$ agrees) and the children as different, so count the solutions of $x_{1} + x_{2} + x_{3} + x_{4} = 16$ with every $x_{i} \geq 1$ .
$^{n - 1} C_{r - 1} =^{15} C_{3} = 455$

Answer key (D) 455

PYQ 7: Integer solutions with an upper limit

JEE Advanced 2026 Paper 1, Mathematics Section 3 Q3

The number of ways to distribute $10$ identical red pens and $14$ identical blue pens among four persons such that each person gets $6$ pens, is _____________.

Solution:

Each person gets $6$ pens, so once the red pens are given out the blue ones are fixed: person $i$ gets $r_{i}$ red and $6 - r_{i}$ blue pens.
So count $r_{1} + r_{2} + r_{3} + r_{4} = 10$ with $0 \leq r_{i} \leq 6$ . The blue pens then add up to $24 - 10 = 14$ by themselves.
Without the upper limit: $^{13} C_{3} = 286$ solutions. With $r_{1} \geq 7$ , put $r_{1} = 7 + t$ : then $t + r_{2} + r_{3} + r_{4} = 3$ has $^{6} C_{3} = 20$ solutions. The same holds for each of the $4$ people, and two people cannot both have $7$ or more.
Number of ways: $286 - 4 \cdot 20 = 206$ .

Answer key 206

PYQ 8: Different objects, no box empty

JEE Main 2026 Apr 8 Shift 2, Mathematics Q9

A person has three different bags and four different books. The number of ways, in which he can put these books in the bags so that no bag is empty, is

(A) $18$
(B) $36$
(C) $39$
(D) $72$

Solution:

Each book goes into one of the $3$ bags: $3^{4} = 81$ ways.
Take away the ways with a given bag empty, $3 \cdot 2^{4} = 48$ , and add back the ways with two given bags empty (taken away twice), $3 \cdot 1^{4} = 3$ .
No bag empty: $81 - 48 + 3 = 36$ . Directly: two books share a bag ( $^{4} C_{2} = 6$ ways), then the three groups go to the three bags in $3! = 6$ ways: $36$ .

Answer key (B) $36$

PYQ 9: Groups, with inclusion-exclusion

JEE Advanced 2024 Paper 1, Mathematics Section 3 Q4

A group of 9 students, $s_{1}, s_{2}, \dots, s_{9}$ , is to be divided to form three teams $X$ , $Y$ and $Z$ of sizes 2, 3, and 4, respectively. Suppose that $s_{1}$ cannot be selected for the team $X$ , and $s_{2}$ cannot be selected for the team $Y$ . Then the number of ways to form such teams, is ______.

Solution:

Without restrictions: $\frac{9 !}{2 ! 3 ! 4 !} = 1260$ ways (the teams are named, so nothing more to divide).
$s_{1}$ in X: one more member for X from the other $8$ , then $3$ of the remaining $7$ for Y: $8 \cdot^{7} C_{3} = 280$ .
$s_{2}$ in Y: two more members for Y from $8$ , then $2$ of the remaining $6$ for X: $^{8} C_{2} \cdot^{6} C_{2} = 420$ .
Both: one more for X from $7$ , then two more for Y from the remaining $6$ : $7 \cdot^{6} C_{2} = 105$ .
Allowed: $1260 - 280 - 420 + 105 = 665$ .

Answer key 665

PYQ 10: Power of a prime in

n!

JEE Main 2026 Jan 21 Shift 2, Mathematics Q20

The largest $n \in N$ , for which $7^{n}$ divides $101!$ , is :

(A) 16
(B) 18
(C) 15
(D) 19

Solution:

$[\frac{101}{7}] + [\frac{101}{49}] = 14 + 2 = 16$
and the sum stops there, since $7^{3} = 343 > 101$ .
So the largest $n$ is $16$ .

Answer key (A) 16

Every formula on this page is checked with a computer algebra system, and every worked answer is recomputed and matched with the answer key. Past papers: JEE Main 2025-2026, JEE Advanced 2022-2026; the questions were read by hand (lists checked 2026-10-15).

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Which formulas JEE uses most

1. Counting principles, nPr​ and nCr​

2. Arrangements

3. Selections, points and lines

4. Distributions, integer solutions and functions

5. Exam traps

6. Past-paper questions, solved

Learn each topic in depth

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Practise questions like these

1. Counting principles, $^{n} P_{r}$ and $^{n} C_{r}$