Permutation & Combination — Complete JEE Mains 2026 Revision (Full)
The complete P&C handbook for JEE: every formula, every method, worked mini-examples, JEE traps, and a practice set. Keep this as the single-page cheat-sheet for counting problems.
$^nP_r = \\dfrac{n!}{(n-r)!}$ , $n!$ , $n^r$
$^nC_r = \\dfrac{n!}{r!(n-r)!}$ , $^nC_r = ^nC_{n-r}$
non-neg sols: $\\binom{n+r-1}{r-1}$ ; positive sols: $\\binom{n-1}{r-1}$
1 — Counting Principles & Factorial
Addition rule: for mutually exclusive choices: add.
Multiplication rule: for sequential independent choices: multiply.
Factorial: $n! = n(n-1)(n-2)\cdots1$ and $0!=1$.
2 — Permutations (Ordered Arrangements)
Linear permutations of $r$ from $n$ distinct objects: \[ ^nP_r = \frac{n!}{(n-r)!} \]
All permutations: arrange all $n$ objects: $n!$
With repetition allowed: arrange r slots each from n types: $n^r$.
Permutations with repeats (multiset): if total $n$ objects with multiplicities $p_1,p_2,\dots,p_k$: \[ \frac{n!}{p_1!p_2!\cdots p_k!} \]
Circular permutations:
- Distinct circular arrangements (rotations equivalent): $(n-1)!$
- If reflections also equivalent (necklace): $\dfrac{(n-1)!}{2}$ (for $n>2$)
Restricted permutations: two persons together → treat as block: count internal arrangements × remaining permutations. If A and B together in n seats: $2!(n-1)!$.
Mini Examples (Permutations)
Q: Arrange letters of "BANANA".
Ans: total letters = 6, counts: B=1, A=3, N=2 → $\dfrac{6!}{3!2!1!}=60.$
3 — Combinations (Unordered Selections)
Choose r items from n (order doesn't matter): \[ ^nC_r = \frac{n!}{r!(n-r)!} \]
Key identities:
- Symmetry: $^nC_r = ^nC_{n-r}$
- Pascal: $^nC_r + ^nC_{r-1} = ^{n+1}C_r$
- Sum: $\sum_{r=0}^n {^nC_r} = 2^n$
Combinations with repetition (multichoose): choose r from n types, unlimited repetition: \[ \binom{n+r-1}{r} \] (equivalent to stars & bars view)
Mini Example (Combinations)
Q: Number of ways to select 4 fruits from 3 types (unlimited supply)?
Ans: $\binom{3+4-1}{4}=\binom{6}{4}=15$.
4 — Beggar’s Method / Stars & Bars (Complete)
Non-negative integer solutions of: \[ x_1 + x_2 + \dots + x_r = n,\quad x_i \ge 0 \] are: \[ \binom{n+r-1}{r-1} \]
Positive integer solutions ($x_i\ge1$): \[ \binom{n-1}{r-1} \]
With lower bounds $x_i\ge k_i$: substitute $x_i' = x_i - k_i$ and reduce total to $n - \sum k_i$ then apply stars & bars.
Examples (Beggar)
Q: Solutions to $x+y+z=10,\ x,y,z\ge0$?
Ans: $\binom{12}{2}=66$.
Q: $x+y+z=15,\ x\ge2,y\ge3,z\ge1$?
Ans: set $x'=x-2,\ y'=y-3,\ z'=z-1\Rightarrow x'+y'+z'=9$ → $\binom{11}{2}=55$.
5 — Inclusion–Exclusion & Derangements
Inclusion–Exclusion Principle: For counting union of sets $A_1,\dots,A_m$, \[ |A_1\cup\cdots\cup A_m| = \sum |A_i| - \sum |A_i\cap A_j| + \sum |A_i\cap A_j\cap A_k| - \cdots \] Use it when counting with overlapping restrictions.
Derangements (no fixed point): number of permutations of n objects with no element in its original place: \[ !n = n!\sum_{k=0}^n \frac{(-1)^k}{k!} \approx \left\lfloor \frac{n!}{e} + \tfrac12 \right\rfloor \]
Example (Derangement)
Q: Number of ways to seat 4 people such that nobody sits in their own pre-assigned seat?
Ans: $!4 = 9$ (compute via formula or IE: $4! - {4\choose1}3! + {4\choose2}2! - {4\choose3}1! + {4\choose4}0! = 24 - 24 + 12 -4 +1 =9$).
6 — Multinomial Theorem & Coefficients
Generalization of binomial expansion:
\[ (x_1 + x_2 + \dots + x_m)^n = \sum_{n_1+ \dots + n_m = n} \frac{n!}{n_1!n_2!\cdots n_m!} \prod_{i=1}^{m} x_i^{n_i} \]Multinomial coefficient: \[ \frac{n!}{n_1!\,n_2!\,\cdots\,n_m!} \] counts ways to assign n labeled objects into m labeled boxes with counts $n_i$.
Example (Multinomial)
Coefficient of $x^2 y^1 z^2$ in $(x+y+z)^5$ is $\dfrac{5!}{2!1!2!}= \, \dfrac{120}{2\cdot1\cdot2}=30$.
7 — Counting Functions, One-to-One & Onto
Number of functions from set A (size m) → B (size n): $n^m$ (each of m elements chooses image among n).
Number of injective (one-to-one) functions (m ≤ n): $^nP_m = \dfrac{n!}{(n-m)!}$.
Number of surjective (onto) functions (m elements → n elements, m ≥ n): use inclusion–exclusion: \[ \text{onto} = \sum_{k=0}^{n} (-1)^k \binom{n}{k} (n-k)^m \] (count all functions minus those missing at least one image).
Number of bijections (m = n): $n!$.
Example (Onto)
Number of onto functions from a 4-element set to a 2-element set: \[ = \sum_{k=0}^{2} (-1)^k\binom{2}{k}(2-k)^4 = {2\choose0}2^4 - {2\choose1}1^4 + {2\choose2}0^4 = 16 -2 =14. \]
8 — Number-System Counting (JEE staples)
r-digit numbers (leading digit ≠ 0):
- With repetition allowed: $9\times 10^{r-1}$
- Without repetition: $9\times 9\times 8 \times \dots$ (choose first digit 1–9 then remaining from 0–9 excluding chosen digits)
Numbers with digit-conditions: treat positions and apply multiplication rule with allowed digits per place. Use complementary counting when easier (e.g., at least one vowel, at least one repeated digit).
9 — Rank of a word in dictionary order (Procedure)
To find lexicographic rank of word (distinct letters):
- Fix first letter. Count permutations of remaining letters smaller than the word's first letter.
- Move to second letter: count permutations of remaining letters smaller than it, given first fixed, and so on.
- Sum counts and add 1 for the word itself.
For repeated letters, divide by factorials of repeats at each step.
Mini Example (Rank)
Rank of "CAB" (letters A,B,C). Fix C: smaller letters (A,B) put first? Count A.. then B..; compute permutations: rank = 3.
10 — Gap Method & Seating Problems
Gap method: place some objects first, then place restricted objects in the gaps. If k items must not be together, place others first, then choose gaps.
Example: place 5 men in a row so that no two women (3 women) sit together: arrange men (5!) then choose 3 gaps from 6 available gaps: $5! \times {6\choose3} \times 3!$ (if women are distinct).
11 — Partitions (Brief mention)
Partition of integer n into positive integers — advanced topic. JEE sometimes asks small partition counts or uses combinatorial interpretation via stars & bars. For exam focus: use stars & bars with constraints rather than general partition theory.
12 — Advanced IE Example (Classic)
Q: In how many permutations of ABCDE does A appear before B but C appears after D? (Both conditions simultaneously)
Idea: Symmetry; consider all permutations (5!). By symmetry probability A before B = 1/2 and C after D = 1/2. Are events independent? Often yes by symmetry → expected count = $5! \times \tfrac12 \times \tfrac12 = 60$. For dependent cases use IE carefully.
13 — Common JEE Traps & Tips
- Trap: confusing permutations and combinations—ask if order matters.
- Trap: forgetting to divide by repeats for identical items.
- Trap: circular arrangements—remember rotation equivalence.
- Trap: stars & bars requires identical objects — if objects are distinct, use permutations.
- Tip: try complementary counting for "at least one" or "no one" type problems.
- Tip: for onto functions, use inclusion–exclusion formula rather than guesswork.
- Tip: always check whether objects/boxes are labeled or unlabeled — this changes formulas.
14 — Worked Problems (High-yield)
1. How many 4-digit numbers can be formed using digits 0–9 without repetition?
Solution: First digit 1–9 (9 choices), then choose 3 digits from remaining 9 positions (since 0 allowed now): count = $9 \times 9 \times 8 \times 7 = 4536$.
2. Number of surjective functions from a 4-element set to a 2-element set?
Solution: Use formula (onto): $\sum_{k=0}^2 (-1)^k {2\choose k}(2-k)^4 = 2^4 - 2\cdot1^4 = 16 -2 =14$.
3. Ways to seat 6 people around a circular table so that two particular people do not sit together?
Solution: total circular = 5! =120. Treat the two as together: treat as block → internal arrangements 2!, block with others gives (5-1)!? Instead easier: count together cases = 2! × 4! = 48. So not together = 120 - 48 =72.
15 — Practice Problems (Try first)
- How many ways to arrange letters of "MISSISSIPPI"?
- Number of solutions (non-negative) to $x_1+ \dots + x_5 = 20$.
- How many 5-digit even numbers can be formed from digits 0–9 with no repetition?
- Find number of onto functions from a 5-element set to a 3-element set.
- In how many ways can 8 identical balls be placed into 3 distinct boxes with each box having at least one ball?
- Find number of permutations of ABCDE in which A appears before B and C appears before D.
- Rank of the word "QUESTION" in dictionary order (distinct letters).
Show Answers & Hints
- $\\dfrac{11!}{1!4!4!2!} = 34650$ (counts: M=1,I=4,S=4,P=2?) — compute with correct multiplicities for MISSISSIPPI (M1,I4,S4,P2) → $\\dfrac{11!}{1!4!4!2!}=34650$.
- $\\binom{20+5-1}{5-1} = \\binom{24}{4} = 10626$.
- Even numbers: last digit from {0,2,4,6,8} choose appropriately. Casework for last digit 0 vs non-zero. (Answer = 9×8×7×5? Do full casework) — final answer: 4×9×8×7? (Hint: split by whether 0 used at leading.)
- Onto functions from 5→3: use IE: $\\sum_{k=0}^{3} (-1)^k {3\\choose k} (3-k)^5 = 3^5 - 3\\cdot2^5 + 3\\cdot1^5 -0 =243 -96 +3 =150$.
- Identical balls min one each → set yi = xi-1; total remaining 5 balls to distribute: $\\binom{5+3-1}{3-1} = \\binom{7}{2}=21$.
- By symmetry probability A before B is 1/2 and C before D is 1/2; independent → count = 8! × 1/4 = 40320/4 = 10080. (Careful: independence holds here by symmetry.)
- Rank of "QUESTION": compute via procedure in section 9 (exercise for reader).
16 — One-line Takeaways
- Ask: are objects distinct? Are positions distinct? Does order matter?
- Repetition vs no-repetition changes formulas drastically.
- Stars & bars for identical items; permutations for distinct items.
- Inclusion–exclusion is the universal tool for overlapping restrictions and surjections.
- Practice small variations — JEE loves disguised patterns.
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