How much of JEE Main comes from Permutations and Combinations?

JEEGenie estimates about 3% of JEE Main marks come from Permutations and Combinations, based on recent papers. The share varies from year to year.

Which formulas matter most in Permutations and Combinations?

The Permutations and Combinations sheet on this page has 24 formula entries, starting with Fundamental Principle of Counting (FPC), Permutation, Combination, Permutations with Repetition, Circular Permutations. Each one comes with when to use it and the trap to avoid.

What topics does Permutations and Combinations cover for JEE?

Permutations and Combinations covers Fundamental Principle of Counting, Permutations, Combinations, Circular Permutations, Selection and Arrangement, Applications.

How should I revise Permutations and Combinations before the exam?

Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Permutations and Combinations questions on JEEGenie.

How many previous-year questions are there on Permutations and Combinations?

JEEGenie has 292 previous-year questions on Permutations and Combinations, and this page shows a few with worked explanations.

Is practice on Permutations and Combinations free on JEEGenie?

Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.

Permutations and Combinations for JEE Main & Advanced — formulas, what is asked, and traps

JEEGenie estimates about 3% of JEE Main marks come from Permutations and Combinations. This page covers 24 formula entries, each with when to use it, what JEE asks and the traps. A common slip: Confusing independent events with mutually exclusive ones; multiplication is for sequences, addition for alternatives.

Key points to know

Fundamental Principle of Counting (FPC). If one event can occur in $m$ ways and another independent event can occur in $n$ ways, then the total number of ways both events can occur in sequence is $m \times n$ (multiplication rule). For mutually exclusive events, add the number of ways (addition rule).
Permutation. A permutation is an arrangement of objects in a specific order. Order matters.
Combination. A combination is a selection of objects where order does not matter.
Factorial Notation. $n!$ is the product of all positive integers from 1 to $n$ , with $0! = 1$ by convention.
Multinomial Theorem for Counting. Number of ways to distribute $n$ distinct objects into $k$ distinct boxes with given counts is $\frac{n!}{n_{1}! n_{2}! \dots n_{k}!}$ .
Derangements (Subfactorial). Number of permutations where no object appears in its original position.
Circular Permutations with Identical Objects. When arranging $n$ objects in a circle with some identical, first fix one object to break symmetry, then arrange the rest considering identical objects.
Binomial Coefficient Properties. Key identities: $(\binom{n}{r}) = (\binom{n}{n - r})$ , $(\binom{n}{r}) + (\binom{n}{r - 1}) = (\binom{n + 1}{r})$ , and $\sum_{r = 0}^{n} (\binom{n}{r}) = 2^{n}$ .

Formula sheet

Name	Formula	When to use
Fundamental Principle of Counting (FPC)	$m \times n$ (multiplication), $m + n$ (addition)	Use when counting total outcomes for multi-step processes or choices from distinct categories.
Permutation	$P (n, r) = \frac{n!}{(n - r)!}$	Use when arranging distinct objects in a sequence, e.g., seating arrangements, word formations, rankings.
Combination	$C (n, r) = (\binom{n}{r}) = \frac{n!}{r! (n - r)!}$	Use when selecting a subset from a set without regard to order, e.g., choosing teams, lottery numbers.
Permutations with Repetition	$\frac{n!}{n_{1}! n_{2}! \dots n_{k}!}$ where $n_{1} + n_{2} + \dots + n_{k} = n$	Use when arranging objects with repetitions, e.g., arranging letters of 'MISSISSIPPI'.
Circular Permutations	$(n - 1)!$ for distinct objects; $\frac{(n - 1)!}{2}$ for arrangements where clockwise and anticlockwise are same (e.g., necklace).	Use for seating around a round table, arranging beads in a necklace, or any cyclic arrangement.
Combinations with Repetition (Stars and Bars)	$(\binom{n + r - 1}{r})$	Use for distributing identical items into distinct boxes, or selecting items with repetition allowed.
Counting with Constraints (Inclusion-Exclusion)	$∣ A \cup B ∣ = ∣ A ∣ + ∣ B ∣ - ∣ A \cap B ∣$ for two sets; extend for more.	Use when counting with 'at least one' or 'none' conditions, e.g., arrangements where no two vowels are together.
Gap Method for Arrangements with Restrictions	Number of gaps = (number of other objects) + 1	Use when arranging objects such that certain items are not adjacent, e.g., no two boys sit together.
Block Method for 'Must Be Together'	Number of ways = (arrangements of block + others) $\times$ (arrangements within block)	Use when certain objects must always be adjacent, e.g., vowels together in a word.
Factorial Notation	$n! = n \times (n - 1) \times \dots \times 1$	Used as building block for all permutation and combination formulas.
Number of Diagonals in a Polygon	$\frac{n (n - 3)}{2}$	Use when counting diagonals in any convex polygon.
Counting with 'At Least' or 'At Most'	Total ways = $\sum_{i = 0}^{k} ways with exactly i$	Use for selection problems like 'at least 2 girls in a committee'.
Multinomial Theorem for Counting	$\frac{n!}{n_{1}! n_{2}! \dots n_{k}!}$ where $\sum n_{i} = n$	Use when distributing distinct items into distinct groups with fixed group sizes.
Number of Ways to Form a Committee with a Chairperson	$(\binom{n}{r}) \times r$ (or $n \times (\binom{n - 1}{r - 1})$ )	Use when a committee has a designated leader or head.
Counting with Identical Objects (Distribution)	$(\binom{n + r - 1}{r - 1})$ for non-negative integers; $(\binom{n - 1}{r - 1})$ for positive integers.	Use for distributing identical candies to children, or solving $x_{1} + . . . + x_{r} = n$ with integer solutions.
Derangements (Subfactorial)	$! n = n! \sum_{k = 0}^{n} \frac{(- 1)^{k}}{k!}$	Use for problems like 'no one gets their own hat back'.
Counting with 'No Two Consecutive'	Number of gaps = (number of unrestricted objects) + 1	Use for arrangements where certain items cannot be adjacent, e.g., no two red balls together.
Number of Ways to Choose with Repetition (Ordered)	$n^{r}$	Use for passwords, PIN codes, or any sequence where each position has same options.
Circular Permutations with Identical Objects	$\frac{(n - 1)!}{n_{1}! n_{2}! \dots n_{k}!}$ valid only when one object is unique (fix it); otherwise use necklace/Burnside counting	Use for arranging beads in a necklace with repeated colors.
Counting with 'At Least One' Using Complement	$Total - ways with none$	Use for 'at least one vowel' or 'at least one girl' in selections.
Number of Ways to Partition a Set into Groups	$\frac{n!}{n_{1}! n_{2}! \dots n_{k}!}$ for labeled groups; divide by $k!$ if unlabeled and equal sizes.	Use for dividing a team into subgroups of specific sizes.
Counting with 'No Two Vowels Together'	Number of gaps = (number of consonants) + 1	Use for word arrangement problems where vowels must not be adjacent.
Binomial Coefficient Properties	$(\binom{n}{r}) = (\binom{n}{n - r})$ , $(\binom{n}{r}) + (\binom{n}{r - 1}) = (\binom{n + 1}{r})$	Use for simplifying combinatorial expressions and in binomial theorem.
Counting with 'Exactly' Conditions	$(\binom{type size}{k}) \times (\binom{others}{r - k})$	Use for selections with specific counts, e.g., exactly 2 boys in a committee.

Open the Permutations and Combinations formula page

What JEE asks

Fundamental Principle of Counting (FPC). JEE often tests FPC in problems like counting number of ways to form numbers, passwords, or selections with constraints.
Permutation. JEE tests permutations with constraints like vowels together, no two consonants adjacent, or circular arrangements.
Combination. JEE often combines combinations with constraints like 'at least one' or 'must include/exclude'.
Permutations with Repetition. JEE tests this with words having repeated letters, or arranging beads in a necklace (circular with identical).
Circular Permutations. JEE often asks number of ways to seat people around a table with constraints like couples sitting together.
Combinations with Repetition (Stars and Bars). JEE tests this in problems like number of solutions to $x_{1} + x_{2} + \dots + x_{n} = r$ with non-negative integers.
Counting with Constraints (Inclusion-Exclusion). JEE often uses inclusion-exclusion for problems like number of ways to arrange letters so that no two vowels are adjacent.
Gap Method for Arrangements with Restrictions. JEE tests this with arrangements of letters where vowels are not together, or seating with gender constraints.
Block Method for 'Must Be Together'. JEE often combines block method with other constraints like no two blocks adjacent.
Factorial Notation. JEE tests factorial simplifications in expressions like $\frac{(2 n)!}{n!}$ .
Number of Diagonals in a Polygon. JEE may ask number of diagonals in a polygon with given number of sides, or intersections of diagonals.
Counting with 'At Least' or 'At Most'. JEE tests this with combinations where certain members must be included or excluded.
Multinomial Theorem for Counting. JEE tests this in problems like number of ways to assign tasks to people with specific numbers.
Number of Ways to Form a Committee with a Chairperson. JEE may ask number of ways to form a team with a captain.
Counting with Identical Objects (Distribution). JEE tests this in problems like number of solutions to equations with constraints.
Derangements (Subfactorial). JEE may ask number of ways to rearrange letters so no letter is in original position.
Counting with 'No Two Consecutive'. JEE tests this with arrangements of letters or seating with constraints.
Number of Ways to Choose with Repetition (Ordered). JEE tests this in problems like number of 3-digit numbers with digits from 1-9 (repetition allowed).
Circular Permutations with Identical Objects. JEE may ask number of distinct necklaces with given beads.
Counting with 'At Least One' Using Complement. JEE tests this in combination problems where certain items must be included.
Number of Ways to Partition a Set into Groups. JEE tests this in problems like number of ways to split 12 students into 3 groups of 4.
Counting with 'No Two Vowels Together'. JEE often tests this with words like 'MATHEMATICS' where letters repeat.
Binomial Coefficient Properties. JEE tests these identities in problems like proving combinatorial sums.
Counting with 'Exactly' Conditions. JEE tests this in combination problems with multiple categories.

Common traps

Confusing independent events with mutually exclusive ones; multiplication is for sequences, addition for alternatives. Fundamental Principle of Counting (FPC)
Forgetting that order matters in permutations; using combinations instead. Permutation
Using permutations when order is irrelevant, e.g., selecting a committee. Combination
Forgetting to account for identical objects; overcounting arrangements. Permutations with Repetition
Forgetting to divide by 2 for arrangements like necklaces where flipping gives same order. Circular Permutations
Using this formula when order matters; use $n^{r}$ instead. Combinations with Repetition (Stars and Bars)
Forgetting to add back intersections when subtracting multiple conditions. Counting with Constraints (Inclusion-Exclusion)
Forgetting to consider gaps at ends; they are valid positions. Gap Method for Arrangements with Restrictions
Forgetting to multiply by internal arrangements of the block. Block Method for 'Must Be Together'
Forgetting that $0! = 1$ , not 0. Factorial Notation
Forgetting to subtract sides; using $(\binom{n}{2})$ directly gives total segments. Number of Diagonals in a Polygon
Forgetting to subtract the case of zero when using complement for 'at least one'. Counting with 'At Least' or 'At Most'
Using this for identical objects; use stars and bars instead. Multinomial Theorem for Counting
Forgetting to multiply by $r$ after selecting committee; chairperson is distinct. Number of Ways to Form a Committee with a Chairperson

Topics in Permutations and Combinations

Fundamental Principle of Counting
Permutations
Combinations
Circular Permutations
Selection and Arrangement
Applications

Previous-year questions on Permutations and Combinations

JEEGenie has 292 previous-year questions on Permutations and Combinations. A few worked examples follow.

Year	Questions
2026	22
2025	22
2024	23
2023	48
2022	24
2021	27
2020	19
2019	18
2018	8
2017	5
2016	7
2015	4
2014	5
2013	6
2012	4
2011	2
2010	1
2009	2
2008	3
2007	2
2006	2
2005	1
2004	4
2003	3
2002	5
2000	1
1998	3
1996	2
1994	1
1992	1
1991	1
1989	1
1988	1
1987	1
1986	1
1985	3
1983	1
1982	3
1981	1
1980	2
1979	2

Question 1 medium

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:

15400
17800
16800
29400

Answer: C

Explanation. We have 5 distinct digits and need 7-digit numbers using each at least once. So exactly two digits are repeated, or one digit appears three times. **Case 1:** One digit appears 3 times, others once. Choose the repeated digit:

(\binom{5}{1}) = 5

ways. Arrange:

\frac{7!}{3!} = 840

ways. Total:

5 \times 840 = 4200

. **Case 2:** Two digits appear twice, three digits once. Choose the two repeated digits:

(\binom{5}{2}) = 10

ways. Arrange:

\frac{7!}{2! 2!} = 1260

ways. Total:

10 \times 1260 = 12600

. Total numbers:

4200 + 12600 = 16800

C

From NTA/IIT papers: JEE Main 2026 · 2 Apr Shift 1

Question 2 easy

The number of sequences of ten terms, whose terms are either 0 or 1 or 2 , that contain exactly five 1 s and exactly three 2 s , is equal to :

360
2520
1820
45

Answer: B

Explanation. We have 10 positions. Choose 5 positions for the 1’s:

(\binom{10}{5})

. From the remaining 5 positions, choose 3 for the 2’s:

(\binom{5}{3})

. The remaining 2 positions automatically get 0’s. Total sequences:

(\binom{10}{5}) \times (\binom{5}{3}) = 252 \times 10 = 2520.

Thus the answer is option B.

From NTA/IIT papers: JEE Main 2025 · 2 Apr Shift 1

Question 3 hard

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 _______.

Answer: 665

Explanation. We first count the total ways to assign the 9 students to teams of sizes 2, 3, 4 without restrictions:

(\binom{9}{2}) (\binom{7}{3}) (\binom{4}{4}) = 36 \cdot 35 \cdot 1 = 1260.

Now subtract the cases where

s_{1}

is in

X

s_{2}

is in

Y

. **Case 1:**

s_{1}

X

. Then choose 1 more for

X

from the remaining 8:

(\binom{8}{1})

. Then choose 3 for

Y

from the remaining 7:

(\binom{7}{3})

. The rest go to

Z

. Count:

(\binom{8}{1}) (\binom{7}{3}) = 8 \cdot 35 = 280.

**Case 2:**

s_{2}

Y

. Choose 2 for

X

from the remaining 8:

(\binom{8}{2})

. Then choose 2 more for

Y

from the remaining 6:

(\binom{6}{2})

. The rest go to

Z

. Count:

(\binom{8}{2}) (\binom{6}{2}) = 28 \cdot 15 = 420.

**Case 3:** Both

s_{1}

X

and

s_{2}

Y

. Choose 1 more for

X

from the remaining 7:

(\binom{7}{1})

. Choose 2 more for

Y

from the remaining 6:

(\binom{6}{2})

. The rest go to

Z

. Count:

(\binom{7}{1}) (\binom{6}{2}) = 7 \cdot 15 = 105.

By inclusion–exclusion, the valid number is:

1260 - 280 - 420 + 105 = 665.

665

From NTA/IIT papers: JEE Advanced 2024

Question 4 medium

The number of ways of giving 20 distinct oranges to 3 children such that each child gets at least one orange is ___________.

Answer: 3483638676

Explanation. We need to distribute 20 distinct oranges among 3 distinct children, with each child getting at least one orange. **Step 1: Total without restriction** Each orange can go to any of the 3 children independently. So total ways =

3^{20}

. **Step 2: Apply inclusion–exclusion for "at least one each"** Let

A_{i}

be the set of distributions where child

i

gets no orange. We want

∣ all ∣ - ∣ A_{1} \cup A_{2} \cup A_{3} ∣

. By inclusion–exclusion:

∣ A_{1} \cup A_{2} \cup A_{3} ∣ = (\binom{3}{1}) 2^{20} - (\binom{3}{2}) 1^{20} + (\binom{3}{3}) 0^{20}

- If one child gets none, the 20 oranges go to the other 2 children:

2^{20}

ways per child omitted. - If two children get none, all oranges go to the remaining 1 child:

1^{20} = 1

way per pair omitted. - If all three get none, impossible since oranges exist:

0^{20} = 0

. So:

∣ A_{1} \cup A_{2} \cup A_{3} ∣ = 3 \cdot 2^{20} - 3 \cdot 1 + 0

**Step 3: Compute the required number**

N = 3^{20} - [3 \cdot 2^{20} - 3]

N = 3^{20} - 3 \cdot 2^{20} + 3

**Step 4: Calculate**

3^{20} = 34867844012^{20} = 10485763 \cdot 2^{20} = 3145728

So:

N = 3486784401 - 3145728 + 3

= 3483638673 + 3

= 3483638676

**Step 5: Final answer**

3483638676.0

From NTA/IIT papers: JEE Main 2023 · 6 Apr Shift 1

Question 5 hard

The number of 4-digit integers in the closed interval

[2022, 4482]

formed by using the digits

0, 2, 3, 4, 6, 7

is ______.

Answer: 569

Explanation. **Principle:** Count digit-strings from the allowed set

{0, 2, 3, 4, 6, 7}

that form 4-digit numbers in

[2022, 4482]

, splitting by leading digit. **Leading digit 2:** Number is

2 a b c

. Need

2 a b c \geq 2022

, so

(a, b, c) \geq (0, 2, 2)

lexicographically. Total with

a, b, c \in {0, 2, 3, 4, 6, 7}

6^{3} = 216

. Subtract those below

022

a = 0

and

b c < 22

, i.e.

b = 0

(6 choices for

c

) or

b = 2, c = 0

(1 choice):

7

. Count

= 216 - 7 = 209

. **Leading digit 3:** All

3 a b c

lie in range. Count

= 6^{3} = 216

. **Leading digit 4:** Need

4 a b c \leq 4482

. If

a < 4

a \in {0, 2, 3}

, giving

3 \cdot 36 = 108

. If

a = 4

: need

b c \leq 82

; all

6^{2} = 36

work since max

b c = 77

. If

a = 6, 7

: exceed. Count

= 108 + 36 = 144

. **Leading digit 0,6,7:** Invalid as 4-digit or exceed

4482

. Total:

209 + 216 + 144 = 569

. Answer: 569

From NTA/IIT papers: JEE Advanced 2022

Frequently asked questions

How much of JEE Main comes from Permutations and Combinations?: JEEGenie estimates about 3% of JEE Main marks come from Permutations and Combinations, based on recent papers. The share varies from year to year.
Which formulas matter most in Permutations and Combinations?: The Permutations and Combinations sheet on this page has 24 formula entries, starting with Fundamental Principle of Counting (FPC), Permutation, Combination, Permutations with Repetition, Circular Permutations. Each one comes with when to use it and the trap to avoid.
What topics does Permutations and Combinations cover for JEE?: Permutations and Combinations covers Fundamental Principle of Counting, Permutations, Combinations, Circular Permutations, Selection and Arrangement, Applications.
How should I revise Permutations and Combinations before the exam?: Learn each formula together with its condition of use, read the trap listed beside it, then practise questions and review every mistake. The practice link on this page opens Permutations and Combinations questions on JEEGenie.
How many previous-year questions are there on Permutations and Combinations?: JEEGenie has 292 previous-year questions on Permutations and Combinations, and this page shows a few with worked explanations.
Is practice on Permutations and Combinations free on JEEGenie?: Sign-up is free and includes a 7-day Pro trial. After the trial, free accounts keep a limited daily quota.