1. Chapter at a Glance
Permutations and combinations are about counting without listing. A permutation is an arrangement (order matters); a combination is a selection (order does not matter). This chapter also powers the Binomial Theorem and Probability, so a strong base here pays off in three chapters. JEE Main asks about arrangements of words, restricted arrangements, selections with conditions, distributions and geometry-based counting.
- Arrangements of letters of a word with repeated letters, and the rank of a word in a dictionary.
- Arrangements with conditions: some objects together, some never together.
- Selections with "at least" or "at most" conditions.
- Number of integer solutions of x1 + x2 + … + xr = n.
2. Fundamental Principles of Counting
- Multiplication principle (AND): if one job can be done in m ways and, after that, another in n ways, both together can be done in m × n ways.
- Addition principle (OR): if two jobs are mutually exclusive and can be done in m and n ways, either one can be done in m + n ways.
Factorial: n! = 1 × 2 × … × n, with 0! = 1.
3. Permutations
This is the number of arrangements of r objects chosen from n distinct objects.
Standard results
- Arrangements of n objects with repetition allowed in r places: nr.
- Arrangements of n objects where p are alike of one kind, q alike of another, and so on: n!/(p! q! …).
- Objects together: tie them into one bundle, arrange the bundles, then arrange inside the bundle.
- Objects never together: arrange the other objects first, then place the restricted objects in the gaps (the "gap method"). With m other objects there are m + 1 gaps.
- Circular arrangements of n distinct objects: (n − 1)!. If clockwise and anticlockwise are the same (necklace, garland): (n − 1)!/2.
Rank of a word in the dictionary
Arrange the letters alphabetically. For each position from the left, count how many unused letters are smaller than the letter actually placed there, and multiply by the factorial of the remaining places. Add all these counts and then add 1.
4. Combinations
- ⁿCr = ⁿCn−r; if ⁿCx = ⁿCy then x = y or x + y = n.
- ⁿCr + ⁿCr−1 = n+1Cr (Pascal's rule).
- ⁿCr is greatest at r = n/2 (n even) or r = (n ± 1)/2 (n odd).
- Selecting at least one from n distinct objects: 2ⁿ − 1.
- From p alike, q alike and r alike objects, selecting at least one: (p + 1)(q + 1)(r + 1) − 1.
Geometry counting
- Lines through n points (no three collinear): ⁿC2; triangles: ⁿC3.
- If m of the n points are collinear: lines = ⁿC2 − ᵐC2 + 1; triangles = ⁿC3 − ᵐC3.
- Diagonals of an n-gon: ⁿC2 − n = n(n − 3)/2.
- Rectangles on an m × n grid of squares: m+1C2 × n+1C2.
5. Distribution and Integer Solutions
| Situation | Number of ways |
|---|---|
| n identical objects into r groups (empty allowed) = non-negative integer solutions of x1 + … + xr = n | n+r−1Cr−1 |
| Each group gets at least one (positive solutions) | n−1Cr−1 |
| n distinct objects into r distinct boxes (empty allowed) | rⁿ |
| Dividing m + n distinct objects into groups of m and n (m ≠ n) | (m + n)!/(m! n!) |
| Dividing 2n distinct objects into two equal groups | (2n)!/(2 (n!)²) |
| Derangements (nobody in the correct place) | n! (1 − 1/1! + 1/2! − … + (−1)ⁿ/n!) |
Derangement values to remember: D2 = 1, D3 = 2, D4 = 9, D5 = 44.
6. Formula Master Table
| Result | Formula |
|---|---|
| Permutations | ⁿPr = n!/(n − r)! |
| Combinations | ⁿCr = n!/(r!(n − r)!) |
| Repeated objects | n!/(p! q! r!) |
| Circular arrangement | (n − 1)! |
| Necklace / garland | (n − 1)!/2 |
| Pascal's rule | ⁿCr + ⁿCr−1 = n+1Cr |
| Sum of combinations | ⁿC0 + ⁿC1 + … + ⁿCn = 2ⁿ |
| Diagonals of n-gon | n(n − 3)/2 |
| Non-negative solutions | n+r−1Cr−1 |
| Positive solutions | n−1Cr−1 |
| Divisors of N = paqbrc | (a + 1)(b + 1)(c + 1) |
7. Mind Map
- AND → multiply
- OR → add
- Complement counting
- ⁿPr, repeated letters
- Together: bundle
- Never together: gaps
- Circular (n − 1)!
- ⁿCr, at least / at most
- Geometry: lines, triangles, diagonals
- Identical objects: stars and bars
- Distinct objects: rⁿ
- Derangements
8. Tips and Tricks
- "At least one" → use the complement: total − none.
- "Never together" = total − together, only when exactly two objects are restricted. For three or more objects that must be separated, use the gap method.
- For vowels together in a word, treat all vowels as one letter; do not forget to arrange the vowels inside the bundle (and divide by repeats inside it).
- Integer-solution problems with lower bounds (x ≥ 2): substitute x = y + 2 to shift to non-negative variables.
- Number of rectangles on an 8 × 8 chessboard = ⁹C2 × ⁹C2 = 1296; number of squares = 1² + 2² + … + 8² = 204.
- Using permutations for "selecting a team" or combinations for "arranging in a row".
- Dividing by 2 in circular arrangements of people — that is only for necklaces or garlands, where flipping gives the same arrangement.
- Double counting in "at least 2 women" problems by first choosing 2 women and then "any" others — always split into cases.
9. Important One-liners
- 0! = 1 and ⁿC0 = ⁿCn = 1.
- ⁿPr = r! × ⁿCr.
- The number of ways to arrange n distinct objects in a row is n!.
- The number of subsets of a set of n elements is 2ⁿ.
- The number of handshakes among n people is ⁿC2.
- The number of ways of selecting r objects from n when p particular objects are always included is n−pCr−p.
- In a knockout tournament with n players, n − 1 matches are played.
- The exponent of a prime p in n! is [n/p] + [n/p²] + [n/p³] + …
10. PYQ-pattern Questions with Solutions
These questions are original and modelled on concepts frequently tested in JEE Main.
In how many ways can the letters of the word MATHEMATICS be arranged?
Show solution
MATHEMATICS has 11 letters: M, A and T each appear twice; H, E, I, C, S once.
Number of arrangements = 11!/(2! 2! 2!) = 39916800/8 = 4989600.
Answer: 4989600
5 boys and 4 girls are to stand in a row so that no two girls are together. The number of ways is:
- 43200
- 2880
- 17280
- 86400
Show solution
Arrange the 5 boys: 5! = 120 ways. This creates 6 gaps (including both ends). Choose and arrange 4 of these gaps for the girls: ⁶P4 = 360.
Total = 120 × 360 = 43200.
Answer: (A)
In how many ways can 8 people sit around a round table if two particular people must not sit next to each other?
- 5040
- 1440
- 3600
- 4320
Show solution
All circular arrangements: (8 − 1)! = 5040. With the two people together: treat them as one unit → 7 units in a circle: 6! = 720, times 2 for their internal order = 1440.
Not together = 5040 − 1440 = 3600.
Answer: (C)
Find the number of non-negative integer solutions of x + y + z = 10.
Show solution
Using the stars-and-bars formula with n = 10 and r = 3: 10+3−1C3−1 = ¹²C2 = 66.
Answer: 66
A committee of 6 is to be formed from 7 men and 5 women so that it has at least 3 women. The number of ways is:
- 462
- 350
- 924
- 455
Show solution
Split into cases by the number of women:
3 women, 3 men: ⁵C3 × ⁷C3 = 10 × 35 = 350.
4 women, 2 men: ⁵C4 × ⁷C2 = 5 × 21 = 105.
5 women, 1 man: ⁵C5 × ⁷C1 = 7.
Total = 350 + 105 + 7 = 462.
Answer: (A)
If all the words formed using all the letters of PENCIL (without repetition) are arranged as in a dictionary, the rank of the word PENCIL is:
- 642
- 643
- 624
- 600
Show solution
Alphabetical order of the letters: C, E, I, L, N, P.
1st letter P: letters smaller than P are C, E, I, L, N (5) → 5 × 5! = 600 words before.
2nd letter E (remaining C, E, I, L, N): smaller is C (1) → 1 × 4! = 24.
3rd letter N (remaining C, I, L, N): smaller are C, I, L (3) → 3 × 3! = 18.
4th letter C (remaining C, I, L): none smaller → 0. 5th letter I (remaining I, L): none → 0.
Words before PENCIL = 600 + 24 + 18 = 642, so the rank is 643.
Answer: (B) 643
How many rectangles (including squares) are there on an 8 × 8 chessboard?
Show solution
A rectangle is fixed by choosing 2 of the 9 horizontal lines and 2 of the 9 vertical lines.
Number = ⁹C2 × ⁹C2 = 36 × 36 = 1296.
Answer: 1296
Out of 12 points in a plane, 5 are collinear and no other three are collinear. The number of triangles that can be formed is:
- 220
- 210
- 200
- 215
Show solution
Without the restriction: ¹²C3 = 220. Three points chosen from the 5 collinear points do not form a triangle: ⁵C3 = 10.
Triangles = 220 − 10 = 210.
Answer: (B)
11. Quick Revision Summary
- Order matters → ⁿPr; order does not matter → ⁿCr.
- Repeated letters: divide by factorials of repeats; circular: (n − 1)!.
- Together → bundle; never together → gap method.
- Identical objects into r groups: n+r−1Cr−1 (at least one each: n−1Cr−1).
- Collinear points reduce lines and triangles; diagonals = n(n − 3)/2.
- "At least" problems: split into cases or use total − complement.