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JEE Mains Maths · Formula sheet

Permutations and Combinations formulas

20 formulas and 20 common traps for JEE Mains Maths Permutations and Combinations, grouped by subtopic.

Full notes with worked examples

Arrangements with Restrictions

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Together, apart and around a table

No two of k items together (row)

n!×n+1Pk(n others arranged, then k items in the gaps)n!\times{}^{n+1}P_{k}\quad(n\ \text{others arranged, then}\ k\ \text{items in the gaps})

Repeated items, sequences and derangements

Arrangements with repeats

n!p! q! r!⋯\frac{n!}{p!\,q!\,r!\cdots}

Common traps

Not together is not the same as apart

'All the vowels never together' allows two of them to be adjacent; it is the total minus the all-together case. 'No two vowels together' is the gap method. Read which one is asked.

People are distinct, floors are distinct

When people leave a lift, each person picks a floor; the order of choosing does not matter but the people do. '4 get off at one floor and 5 at another' is a choice of the 4 people times an ordered pair of floors.

Dictionary Order and Ranks

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The rank of a word

Rank

rank=1+∑i=1nci (n−i)!(ci=smaller unused letters at step i)\text{rank}=1+\sum_{i=1}^{n}c_i\,(n-i)!\quad(c_i=\text{smaller unused letters at step }i)

Numbers listed in order

Block size with repetition

numbers after fixing k of n digits=d n−k\text{numbers after fixing }k\text{ of }n\text{ digits}=d^{\,n-k}

Common traps

Repeated letters change the block size

With a letter still repeated among the unused ones, a block has (n−k)!p!\frac{(n-k)!}{p!} words, not (n−k)!(n-k)!. Recompute the divisor at each step, since using up one copy of a repeated letter changes it.

Zero is allowed after the first digit

When 0 is among the digits, it cannot lead but can appear anywhere else. Blocks for the first digit use d−1d-1 choices; later blocks use all dd.

Forming Numbers from Digits

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Filling positions: ranges, leading digits and no repetition

Sum of all arrangements

∑=(number of arrangements)×(digit sum)n×11⋯1⏟n\sum=\frac{(\text{number of arrangements})\times(\text{digit sum})}{n}\times\underbrace{11\cdots1}_{n}

Divisibility conditions

Digit-sum test

3∣d1d2⋯dn‾ exactly when 3∣d1+d2+⋯+dn3\mid\overline{d_1d_2\cdots d_n}\ \text{exactly when}\ 3\mid d_1+d_2+\cdots+d_n

Fixed digit sum or product

Stars and bars

x1+⋯+xk=S, xi≥0: (S+k−1k−1)x_1+\cdots+x_k=S,\ x_i\ge0:\ \binom{S+k-1}{k-1}

Common traps

Zero in the leading place

When 0 is available, the leading digit has one fewer choice. If the last digit is also restricted (say even, and 0 is even), split into cases 'last digit 0' and 'last digit not 0'.

Count the complement when it is smaller

'Not divisible by 3' is usually easier as total minus divisible. Listing the digit triples whose sum is a multiple of 3 is short; listing the others is long.

Digits stop at 9

Stars and bars counts solutions with any size of digit. For a sum above 9, remove the solutions where some digit is 10 or more before answering.

Selections and Committees

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Committees and 'at least' conditions

Case sum

∑allowed (a,b)(ma)(nb)\sum_{\text{allowed }(a,b)}\binom ma\binom nb

Choosing letters from a word

One pattern

(choices of letters)×r!(repeats)!\text{(choices of letters)}\times\frac{r!}{\text{(repeats)}!}

Working with the nCr and nPr formulas

Telescoping

∑k=1nk⋅k!=(n+1)!−1\sum_{k=1}^{n}k\cdot k!=(n+1)!-1

Common traps

Do not choose 'at least one' first

Choosing one woman first and then any others counts the same committee several times. Split into exact cases, or use the complement.

Distinct letters, not letter count

UNIVERSE has 8 letters but E repeats; without repetition there are only 3 distinct vowels. Count distinct letters before choosing.

Check which root is valid

Equations in nn from these formulas are often quadratic. Discard a root that is negative, not an integer, or smaller than rr.

Distributions and Integer Solutions

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Identical objects: stars and bars

Stars and bars

x1+⋯+xk=n, xi≥0:(n+k−1k−1)x_1+\cdots+x_k=n,\ x_i\ge0:\quad\binom{n+k-1}{k-1}

Distinct objects: onto maps and groups

Onto maps

∑j=0k(−1)j(kj)(k−j)n\sum_{j=0}^{k}(-1)^j\binom kj(k-j)^n

Common traps

Distinct values are a separate count

'x,y,zx,y,z distinct' is not built into stars and bars. Count all solutions, then subtract those with two or three equal values, using inclusion–exclusion.

Labelled or unlabelled

Cars of different makes are labelled, so which car gets 2 people matters. Unnamed groups of equal size are not: divide by the ways to permute those groups.

Counting Functions, Matrices and Subsets

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Counting functions

One-one and increasing

#{one-one}=nPm,#{strictly increasing}=(nm)\#\{\text{one-one}\}={}^nP_m,\qquad\#\{\text{strictly increasing}\}=\binom nm

Counting matrices by their entries

Trace of AᵀA

tr⁡(ATA)=∑i,jaij2\operatorname{tr}(A^{T}A)=\sum_{i,j}a_{ij}^{2}

Subsets with a property

Complement

#{contains an even}=2n−2#odd\#\{\text{contains an even}\}=2^n-2^{\#\text{odd}}

Common traps

Constrained inputs first

Assigning free inputs first can use up a value a constrained input needs, and the count then depends on earlier choices. Fix the constrained values first so every later step has a fixed number of options.

Signs multiply only non-zero entries

A zero entry has one form; each non-zero entry from {−2,−1,1,2}\{-2,-1,1,2\} has two signs. Multiply by 2(number of non-zero entries)2^{(\text{number of non-zero entries})}, not 2(cells)2^{(\text{cells})}.

The empty set

Check whether the question counts the empty set. Its sum is 0 — a multiple of 3 — and its product is taken as 1; 'non-empty' removes it.

Points, Lines and Polygons

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Collinear points, parallel and concurrent lines

Triangles from points with collinear sets

(n3)−∑i(ki3)\binom n3-\sum_i\binom{k_i}{3}

Triangles and diagonals in a polygon

No side of the polygon

n(n−4)(n−5)6\frac{n(n-4)(n-5)}{6}

Common traps

Vertices are not among the points

When the points lie in the interior of the sides, the triangle's own vertices are not available. Use only the given points, and subtract collinear triples side by side.

Indices on a circle

For points P1,…,PnP_1,\dots,P_n on a circle, any three give a triangle; a condition on the indices (like i+j+k≠15i+j+k\neq15) removes only the listed triples. List them carefully, with distinct indices.

Divisibility, Divisors and Factorials

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Powers of a prime in n!, and counting divisors

Legendre's formula

vp(n!)=⌊np⌋+⌊np2⌋+⌊np3⌋+⋯v_p(n!)=\left\lfloor\frac np\right\rfloor+\left\lfloor\frac n{p^2}\right\rfloor+\left\lfloor\frac n{p^3}\right\rfloor+\cdots

Counting multiples in a range

Inclusion–exclusion

∣A∪B∣=∣A∣+∣B∣−∣A∩B∣|A\cup B|=|A|+|B|-|A\cap B|

Pairing numbers by their remainders

Matching classes

#{x+y≡0}=∑r+s≡0∣Cr∣ ∣Cs∣\#\{x+y\equiv0\}=\sum_{r+s\equiv0}|C_r|\,|C_s|

Common traps

The scarcer prime decides

For 40n∣n!40^n\mid n! you need both 23n2^{3n} and 5n5^n. Compute each limit and take the smaller; the prime that appears less often usually wins.

Use the lcm, not the product

Numbers divisible by both 4 and 6 are multiples of 12, not of 24. Always take the lcm for the overlap.

Ordered or unordered

'Ways of choosing xx and yy' with named variables counts ordered pairs. Decide before multiplying, and for pairs inside one class use k(k−1)k(k-1) ordered or (k2)\binom k2 unordered.

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