Permutations and Combinations Formula Sheet — JEE Main Mathematics

Every key Permutations and Combinations formula, definition and theorem for JEE Main Mathematics in one place — with common examiner traps and worked examples. Free to read; blurt from memory, then check your gaps.

Syllabus — topics coveredNTA · 5 sub-topics

Fundamental principle of counting
Factorial n
Permutations and combinations
Derivation of formulae and their connections
Simple applications

Counting Principle & Factorials

Fundamental principle of counting:

Multiply

for tasks done in succession ('this AND then that');

add

for mutually exclusive alternatives ('this OR that'). Almost every counting problem is built from these two rules.

★ Remember · AND multiplies, OR adds

Filling several places in a row (each a sub-choice)

\Rightarrow

multiply

the counts. Splitting into separate non-overlapping cases

\Rightarrow

add

the case counts. Getting these two backwards is the most common slip.

🎯 Exam · Factorial essentials

n! = 1 \cdot 2 \cdot 3 \dots n

, with

0! = 1

and

n! = n (n - 1)!

. The ratio

\frac{n !}{( n - r )!} = n (n - 1) \dots (n - r + 1)

is exactly r falling factors — the backbone of

^{n} P_{r}

★ Remember · Highest power of a prime in n!

The exponent of a prime p in n! is

⌊ \frac{n}{p} ⌋ + ⌊ \frac{n}{p ^{2}} ⌋ + \dots

(Legendre's formula) — useful for trailing-zero and divisibility questions about factorials.

🚫 Examiner Trap · Examiner traps

(1) AND

\Rightarrow

MULTIPLY (multiplication principle); OR (mutually exclusive)

\Rightarrow

ADD — mixing these up is the #1 error. (2)

0! = 1

by convention (not

0

). (3) The addition rule needs the cases to be MUTUALLY EXCLUSIVE, else subtract the overlap. (4) Exponent of a prime p in n! is

⌊ n / p ⌋ + ⌊ n / p^{2} ⌋ + \dots

(Legendre) — used for trailing zeros.

Permutations — Arrangements & Repetition

Permutation: An

arrangement

in which order matters. The number of arrangements of n distinct objects taken r at a time (no repetition) is

^{n} P_{r} = \frac{n !}{( n - r )!} = n (n - 1) \dots (n - r + 1)

Key permutation counts

▸All n taken together: $^{n} P_{n} = n!$ .
▸Repetition allowed, r places from n kinds: $n^{r}$ .
▸ $^{n} P_{0} = 1$ and $^{n} P_{1} = n$ .
▸Link to selections: $^{n} P_{r} =^{n} C_{r} \cdot r!$ .

★ Remember · Think in slots

Draw the r blanks. Fill the first in n ways, the next in

n - 1

, and so on (no repetition) — multiplying gives

^{n} P_{r}

. If repetition is allowed, every blank has n choices, giving

n^{r}

⚠️ Watch out · Order matters ⇒ permutation

Use a permutation when rearranging gives a

different

outcome (a ranking, a number, a seating). If swapping two chosen items changes nothing (a committee, a hand of cards), it is a

combination

instead.

🚫 Examiner Trap · Examiner traps

(1) Permutation

=

order MATTERS; combination

=

order doesn't — decide this first. (2)

^{n} P_{r} = \frac{n !}{( n - r )!} =^{n} C_{r} \cdot r!

(a selection

\times

its arrangements). (3) With repetition allowed it's

n^{r}

, NOT

^{n} P_{r}

. (4)

^{n} P_{0} = 1

and

^{n} P_{n} = n!

(no

0!

slip).

Permutations — Restrictions, Alike Objects & Circular

Arrangements with repeated (alike) objects, the together and never-together rules, circular permutations, and the gap and necklace methods — Restricted arrangements, the gap method and circular permutations.

★ Remember · When some objects are alike

If among n objects p are alike of one kind, q of another, ..., the number of distinct arrangements is

\frac{n !}{p ! q ! \dots}

(e.g. the letters of MISSISSIPPI). Divide out the rearrangements you cannot tell apart.

Together / not together / circular

▸ $Together$ : tie the k items into one block $\Rightarrow (n - k + 1)! \cdot k!$ .
▸ $Never together$ : total $-$ together, or the gap method.
▸ $Gap method$ : arrange m others (m!), drop the rest into the $m + 1$ gaps.
▸ $Circular$ of n distinct: $(n - 1)!$ ; necklace/garland: $\frac{( n - 1 )!}{2}$ .

🎯 Exam · Why circular is (n−1)!

On a round table, every arrangement can be rotated into n identical-looking copies, so we fix one person and arrange the remaining

n - 1

(n - 1)!

. If a flip also looks the same (beads on a string), halve it to

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

⚠️ Watch out · Gap method needs the rest placed first

For 'no two of a set together',

first

arrange the unrestricted items,

then

slot the restricted ones into the gaps — never the other way round, or you over-count adjacencies.

🚫 Examiner Trap · Examiner traps

(1) Alike objects DIVIDE: n things with

p, q, \dots

alike

= \frac{n !}{p ! q ! \dots}

. (2) Circular arrangements are

(n - 1)!

(fix one to kill rotation), and

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

for a necklace/garland (reflection counts as same). (3) 'Together'

=

tie into one block

\Rightarrow (n - k + 1)! k!

; 'never together'

=

total

-

together (or GAP method). (4) Gap method: arrange m free items (m!), then drop the rest into the

m + 1

gaps.

Combinations — nCr & Its Identities

Combination: A

selection

where order does not matter. The number of ways to choose r from n distinct objects is

^{n} C_{r} = \frac{n !}{r ! ( n - r )!} = \frac{^{n} P _{r}}{r !}

— a permutation is a selection followed by an arrangement.

Combination identities

Identity	Formula
Symmetry	$^{n} C_{r} =^{n} C_{n - r}$
Pascal	$^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
Row sum	$\sum_{r = 0}^{n}^{n} C_{r} = 2^{n}$
Ratio	$\frac{^{n} C _{r}}{^{n} C _{r - 1}} = \frac{n - r + 1}{r}$
$^{n} C_{x} =^{n} C_{y}$	$\Rightarrow x = y$ or $x + y = n$
Greatest term	$^{n} C_{r}$ largest at $r = ⌊ n /2 ⌋$

Identities to know

▸Symmetry: $^{n} C_{r} =^{n} C_{n - r}$ .
▸Pascal: $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ .
▸Row sum: $^{n} C_{0} +^{n} C_{1} + \dots +^{n} C_{n} = 2^{n}$ .
▸If $^{n} C_{x} =^{n} C_{y}$ then $x = y$ or $x + y = n$ .

🎯 Exam · The ratio trick

\frac{^{n} C _{r}}{^{n} C _{r - 1}} = \frac{n - r + 1}{r}

— the fastest way to find where

^{n} C_{r}

is increasing/decreasing. It is greatest at

r = ⌊ n /2 ⌋

(the middle term).

★ Remember · Selection then arrangement

Many 'word' problems split into two steps: first

choose

the letters (

^{n} C_{r}

), then

arrange

them (multiply by the appropriate permutation count). Keep the two steps separate to avoid double counting.

🚫 Examiner Trap · Examiner traps

(1)

^{n} C_{r} =^{n} C_{n - r}

(symmetry) — use the smaller r to compute. (2)

^{n} C_{x} =^{n} C_{y} \Rightarrow x = y

x + y = n

(don't drop the second case). (3) Pascal:

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

. (4)

^{n} C_{r}

is largest at

r = ⌊ n /2 ⌋

(middle term);

\sum_{r}^{n} C_{r} = 2^{n}

Selection Patterns — Subsets, 'At Least' & Repetition

Subsets and 'at least'

▸Total subsets of n distinct objects: $2^{n}$ (each object in or out).
▸Select at least one: $2^{n} - 1$ (drop the empty set).
▸At least one from groups of $p, q, r$ alike: $(p + 1) (q + 1) (r + 1) - 1$ .
▸Number of divisors of $N = p^{a} q^{b} r^{c}$ : $(a + 1) (b + 1) (c + 1)$ .

🎯 Exam · Selection with repetition

Choosing r objects from n kinds with unlimited repetition (order irrelevant) is

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

— the same stars-and-bars count as putting r identical balls into n boxes.

★ Remember · Complementary counting

For 'at least one' conditions it is almost always faster to count the

total

and subtract the 'none' case:

(at least one) = (total) - (none)

. The

2^{n} - 1

formula is exactly this idea.

⚠️ Watch out · Alike vs distinct changes everything

Selecting from

alike

objects uses the

(p + 1)

'how many to take' factor; selecting from

distinct

objects uses

^{n} C_{r}

2^{n}

. Read carefully whether the objects are identical or not before choosing a formula.

🚫 Examiner Trap · Examiner traps

(1) 'At least one'

=

(total)

-

(none): total subsets

2^{n}

, so at-least-one

= 2^{n} - 1

. (2) Select

\geq 1

from groups of p,q,r ALIKE

= (p + 1) (q + 1) (r + 1) - 1

(the

- 1

removes 'pick nothing'). (3) Choose r with repetition from n kinds

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

. (4) Number of divisors of

N = p^{a} q^{b} r^{c}

(a + 1) (b + 1) (c + 1)

— count exponent choices.

Distribution of Objects into Groups & Boxes

★ Remember · Stars and bars (identical objects)

Distributing n

identical

objects into r

distinct

boxes = non-negative integer solutions of

x_{1} + \dots + x_{r} = n

: each box

\geq 0

gives

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

; each box

\geq 1

gives

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

Distinct objects

▸Into r distinct boxes (any number each): $r^{n}$ .
▸Into labelled groups of sizes p,q,s: $\frac{n !}{p ! q ! s !}$ .
▸If $k$ groups are equal-sized and unlabelled: divide by an extra $k!$ .
▸Onto distributions (no empty box) need inclusion–exclusion.

🎯 Exam · Labelled vs unlabelled groups

52 cards to 4 named players of 13 each:

\frac{52 !}{( 13 ! ) ^{4}}

. The same cards into 4

unnamed

equal piles:

\frac{52 !}{( 13 ! ) ^{4} 4 !}

— divide by

4!

because the piles are indistinguishable.

⚠️ Watch out · Equal groups need the extra k!

Forgetting to divide by k! when k groups have the same size is the classic over-counting error. Always ask: are the groups

distinguishable

(named) or not?

🚫 Examiner Trap · Examiner traps

(1) IDENTICAL objects into distinct boxes (stars & bars):

\geq 0 \Rightarrow^{n + r - 1} C_{r - 1}

\geq 1 \Rightarrow^{n - 1} C_{r - 1}

— pick the right floor. (2) DISTINCT objects into r distinct boxes

= r^{n}

(each object chooses a box). (3) Dividing into groups of sizes p,q,s: labelled

= \frac{n !}{p ! q ! s !}

; if k groups are EQUAL and UNLABELLED, divide by an extra k!. (4) Don't double-count equal unnamed piles.

Applications — Geometry, Word Rank & Derangements

★ Remember · Geometry from n points (no 3 collinear)

Lines

=^{n} C_{2}

, triangles

=^{n} C_{3}

, diagonals of an n-gon

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

, handshakes

=^{n} C_{2}

. If m points are collinear, subtract

^{m} C_{2}

from lines and

^{m} C_{3}

from triangles.

🎯 Exam · Rank of a word

List letters in dictionary order. Scanning left to right, at each position count the unused letters

smaller

than the current one, multiply by the factorial of the remaining places, sum these, and add

1

for the word itself.

★ Remember · Derangements

No object in its original place:

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

, so

D_{1} = 0, D_{2} = 1, D_{3} = 2, D_{4} = 9, D_{5} = 44

(also

D_{n} = (n - 1) (D_{n - 1} + D_{n - 2})

⚠️ Watch out · Watch for repeated letters in rank problems

When a word has repeated letters, the factorial of the remaining places must be divided by the factorials of the repeats (the alike-objects rule) at each step — otherwise the rank comes out too large.

🚫 Examiner Trap · Examiner traps

(1) With n points, no 3 collinear: lines

=^{n} C_{2}

, triangles

=^{n} C_{3}

, diagonals of an n-gon

=^{n} C_{2} - n

. (2) If m points ARE collinear, SUBTRACT

(2 m)

lines (they give 1 not

(2 m)

) and

(3 m)

triangles. (3) Word rank: count smaller unused letters at each position,

\times

factorial of remaining places, sum

+ 1

. (4) Derangements

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

;

D_{1} = 0, D_{2} = 1, D_{3} = 2, D_{4} = 9

More JEE Main Mathematics formula sheets

Frequently Asked Questions

What are the most important Permutations and Combinations formulas for JEE Main?

This Permutations and Combinations formula sheet covers all the high-yield Mathematics formulas, definitions and theorems you need for JEE Main, across Fundamental principle of counting, Factorial n, Permutations and combinations, Derivation of formulae and their connections, Simple applications — each shown with the key result and, where useful, a worked example.

Is this Permutations and Combinations formula sheet free?

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How should I revise Permutations and Combinations formulas?

Blurt the Permutations and Combinations formulas from memory, then check against this sheet to find your gaps — and practise a few previous-year questions on the chapter to make sure you can apply them under time pressure.

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