A complete, exam-focused guide to permutation and combination — clear definitions, the nPr and nCr formulas, the key difference between the two, worked examples, and practice questions for Class 11, Class 12, and JEE Mains.
Combinatorics is the branch of mathematics concerned with counting, arranging, and selecting objects. Permutation and combination are its two most fundamental tools, used to answer one basic question in different forms: "In how many ways can this happen?"
Both appear throughout Class 11 and Class 12 mathematics, in probability, statistics, the binomial theorem, and in competitive exams such as JEE Mains — largely because counting problems show up everywhere, from arranging books on a shelf to forming a committee.
An ordered arrangement of a set of objects. The sequence ABC is different from BCA.
An unordered selection of a set of objects. The group {A, B, C} is the same as {B, C, A}.
Both permutation and combination formulas are built from the factorial function, written n!, which is the product of all positive integers up to n.
By convention, 0! = 1. Example: 5! = 5 × 4 × 3 × 2 × 1 = 120
| n | n! | Value |
|---|---|---|
| 0 | 0! | 1 |
| 1 | 1! | 1 |
| 2 | 2! | 2 |
| 3 | 3! | 6 |
| 4 | 4! | 24 |
| 5 | 5! | 120 |
| 6 | 6! | 720 |
A permutation is an arrangement of objects in a specific order. The number of permutations of r objects chosen from a set of n distinct objects is denoted nPr (read "n permute r").
n = total number of distinct objects | r = number of objects arranged | Valid for 0 ≤ r ≤ n
Figure 1: All 6 permutations of the 3 letters A, B, C. Since order matters, ABC and BCA are counted as two distinct outcomes.
nPr can be understood without the factorial formula: you have n choices for the first position, (n−1) choices for the second (one is used), (n−2) for the third, and so on, for r positions total. Multiplying these choices together gives the same result as n!/(n−r)!.
A combination is a selection of objects where order does not matter. The number of combinations of r objects chosen from n distinct objects is denoted nCr (read "n choose r").
Equivalently: nCr = nPr / r! | This divides out the r! ways each group could be ordered
Figure 2: Only 3 unique combinations exist when choosing 2 letters from 3, since {A,B} and {B,A} represent the same selection.
This is the single most searched question on this topic: what is the difference between permutation and combination? The table below summarizes it clearly.
| Aspect | Permutation | Combination |
|---|---|---|
| Order | Matters (ABC ≠ BCA) | Does not matter (ABC = BCA) |
| Meaning | Arrangement | Selection |
| Formula | nPr = n!/(n−r)! | nCr = n!/[r!(n−r)!] |
| Relationship | nPr = nCr × r! | nCr = nPr ÷ r! |
| Count (same n, r) | Always ≥ combinations | Always ≤ permutations |
| Typical scenario | Ranking, passwords, seating, race positions | Committees, teams, lottery numbers, groups |
| Keyword clues | "arrange," "order," "rank," "sequence," "code" | "select," "choose," "group," "committee," "team" |
Figure 3: nPr and nCr are directly related by r! — every combination expands into r! permutations.
Ask yourself: "If I swap two of the chosen items, do I get a different outcome?" If yes, it's a permutation. If no, it's a combination.
Assigning gold/silver/bronze medals, creating a PIN code, arranging books on a shelf, seating people in specific chairs, ranking finishers in a race.
Choosing a committee of 5 from 20 people, picking lottery numbers, selecting a cricket team, forming a group project, choosing toppings for a pizza.
In how many ways can 4 different books be arranged on a shelf?
SolutionThis is a permutation — order matters on a shelf. Here n = 4, r = 4.
4P4 = 4!/(4−4)! = 4!/0! = 24/1 = 24 ways
In how many ways can a committee of 3 people be chosen from a group of 8?
SolutionThis is a combination — a committee has no internal order. Here n = 8, r = 3.
8C3 = 8!/[3!(8−3)!] = 8!/(3!×5!) = (8×7×6)/(3×2×1) = 336/6 = 56 ways
10 athletes compete in a race. In how many ways can gold, silver, and bronze medals be awarded?
SolutionMedal positions matter — this is a permutation. Here n = 10, r = 3.
10P3 = 10!/(10−3)! = 10!/7! = 10×9×8 = 720 ways
From a standard deck of 52 cards, in how many ways can 5 cards be chosen (a poker hand)?
SolutionA hand of cards has no order — this is a combination. Here n = 52, r = 5.
52C5 = 52!/[5!(47)!] = (52×51×50×49×48)/(5×4×3×2×1) = 2,598,960 ways
| Case | Formula | Note |
|---|---|---|
| Permutations of n objects, all taken | nPn = n! | r = n, so (n−r)! = 0! = 1 |
| Combinations of n objects, all taken | nCn = 1 | Only one way to select everything |
| Combinations, choosing none | nC0 = 1 | One way to select nothing (empty set) |
| Symmetry rule | nCr = nC(n−r) | Choosing r is the same count as choosing what's left out |
| Permutations with repetition allowed | n^r | Each of r positions has n independent choices |
| Permutations of n objects with repeated items | n! / (p!q!...) | p, q = counts of each repeated item, e.g. letters in "MISSISSIPPI" |
| Circular permutations | (n−1)! | Arrangements around a circle, rotations counted as identical |
Permutation and combination are not isolated topics — they are the foundation for two major areas of Class 11–12 mathematics.
Many probability problems are really counting problems: P(event) = favorable outcomes ÷ total outcomes, and both numerator and denominator are usually computed using nCr, since the order in which cards are dealt or dice land typically doesn't matter to the outcome being measured.
The binomial theorem expands (x + y)ⁿ using combinations directly: the coefficient of each term is nCr, called the binomial coefficient. This same idea extends into the binomial distribution in statistics, where nCr counts the number of ways r successes can occur in n independent trials.
Each term's coefficient nCr is exactly the combination formula from this article
| # | Question | Type |
|---|---|---|
| 1 | How many 4-digit numbers can be formed from digits 1–9 with no repetition? | Permutation |
| 2 | In how many ways can 3 students be chosen from a class of 30 for a quiz team? | Combination |
| 3 | How many distinct arrangements exist for the letters of the word "APPLE"? | Permutation with repetition |
| 4 | A bag has 5 red and 4 blue balls. In how many ways can 3 balls be drawn so that at least 2 are red? | Combination |
| 5 | In how many ways can 6 people be seated around a circular table? | Circular permutation |
| 6 | How many ways can a password of 4 distinct digits (0–9) be created? | Permutation |
Permutation and combination are two counting methods in mathematics used to determine the number of ways objects can be arranged or selected from a set. A permutation counts the number of ways to arrange objects where order matters. A combination counts the number of ways to select objects where order does not matter. Both are core topics of combinatorics, the branch of mathematics dealing with counting.
The key difference between permutation and combination is order. A permutation is an ordered arrangement of objects — ABC and BCA are counted as different permutations. A combination is an unordered selection of objects — ABC and BCA are counted as the same combination. Because permutations count every ordering separately, the number of permutations is always greater than or equal to the number of combinations for the same n and r.
The permutation formula is nPr = n!/(n−r)!, which gives the number of ways to arrange r objects from a set of n distinct objects where order matters. The combination formula is nCr = n!/[r!(n−r)!], which gives the number of ways to select r objects from n where order does not matter. Note that nCr = nPr / r!, since each combination corresponds to r! different orderings, all counted separately in nPr.
Use permutation when the arrangement or order matters, such as assigning ranks in a race, forming a password, or seating people in specific chairs. Use combination when only the selection matters and order is irrelevant, such as choosing a committee of people, picking lottery numbers, or selecting a team from a group of players. A simple test: if swapping two chosen items creates a new, distinct outcome, it's a permutation; if it doesn't, it's a combination.
Permutation and combination can feel confusing at first because both involve counting and share similar formulas, but the concept becomes straightforward once you focus on a single question: does order matter? Most difficulty comes from misidentifying which formula applies to a word problem, not from the formulas themselves. Practicing a variety of solved examples is the fastest way to build confidence with this topic.
nPr denotes the number of permutations of r objects chosen from a set of n distinct objects, calculated as n!/(n−r)!. nCr denotes the number of combinations of r objects chosen from n, calculated as n!/[r!(n−r)!]. nPr is read as "n permute r" and nCr is read as "n choose r". Both are fundamental tools in combinatorics and probability calculations.
Combinations are used directly in probability to count favorable outcomes over total outcomes when order doesn't matter, such as in card and dice problems. The binomial theorem also relies on combinations: the coefficient of each term in the expansion of (x + y)^n is given by nCr, known as the binomial coefficient. This connects permutation and combination directly to binomial expansion and binomial distribution in statistics.
