Why is nCr equal to nC(n-r)?

Choosing r objects to include is the same as choosing n-r objects to exclude, so the two selections determine each other and their counts are equal.

Combinations

Q: What is the difference between permutation and combination?

A permutation is an ordered arrangement where order matters; a combination is an unordered selection where order does not matter. nPr = r! times nCr.

Q: When is nCr used instead of nPr?

Use nCr when the question asks to choose, select, or form a group where order does not matter. Use nPr when arranging, seating, ranking, or forming numbers or words.

Q: What is Pascal's identity and why does it matter?

Pascal's identity states nCr + nC(r-1) = (n+1)Cr. When choosing r from n+1 objects, either the last object is included or excluded, giving these two cases. It generates Pascal's triangle.

Q: How do I count triangles with vertices on a polygon?

Any 3 vertices form a triangle so the total is nC3 for an n-sided polygon. For triangles with exactly one side common with the polygon, fix a side (n ways) and pick a third vertex not adjacent to it (n-4 ways), giving n(n-4).

Q: Why do we divide by 3! when splitting 12 books into 3 equal groups?

The three groups are of equal size and unlabelled. Computing 12!/(4! 4! 4!) treats them as labelled, overcounting by 3!. Dividing by 3! corrects this.

Q: How do I find the number of divisors of a number?

Prime-factorise the number as N = p1^a1 times p2^a2 and so on. The number of positive divisors is (a1+1)(a2+1) and so on. Each divisor picks an exponent for each prime independently from 0 to ai.

Q: What does stars and bars mean in JEE?

Stars and bars is a technique to distribute n identical objects into r distinct boxes. Arrange n stars and (r-1) bars in a row; the number of such arrangements is (n+r-1)C(r-1).

Q: Is combinations important for JEE Mains?

Yes. Combinations appear in 1 to 2 direct questions per year in JEE Mains and underpin Probability and Binomial Theorem questions.

MathsPermutations And CombinationsFor JEE aspirants

A combination is an unordered selection of objects from a set. When we pick $r$ objects from a set of $n$ distinct objects without caring about the order, the number of possible selections is $^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$ . Combinations underpin dozens of JEE Mains problems: forming committees, counting handshakes, choosing points to make triangles or diagonals in a polygon, distributing identical objects, and calculating the total number of subsets of a set $(2^{n})$ .

Key Formulas - Quick Reference

Combinations without repetition: $^{n} C_{r} = \frac{n !}{r ! ( n - r )!}$ , where $0 \leq r \leq n$ .
With repetition ( $n$ distinct objects, $r$ selected, each object may repeat): $^{n} H_{r} =^{n + r - 1} C_{r}$ .
Symmetry: $^{n} C_{r} =^{n} C_{n - r}$
Pascal's identity: $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
Ratio identity: $\frac{^{n} C _{r}}{^{n} C _{r - 1}} = \frac{n - r + 1}{r}$
Restricted selection (of $r$ from $n$ ; $k$ particular objects): always included $=^{n - k} C_{r - k}$ ; never included $=^{n - k} C_{r}$ .
Division into groups. Unequal groups of sizes $p, q, r$ (with $p + q + r = n$ ): $\frac{n !}{p ! q ! r !}$ . Equal groups: divide further by the factorial of the number of equal-sized groups.
Identical objects distributed among distinct boxes ( $n$ identical objects, $r$ boxes): $^{n + r - 1} C_{r - 1}$ (each box may be empty); $^{n - 1} C_{r - 1}$ (each box must have at least one).
Total selections from $n$ distinct objects (at least one chosen) $= 2^{n} - 1$ .
Number of divisors of $N = p_{1}^{a_{1}} p_{2}^{a_{2}} \dots p_{k}^{a_{k}}$ is $(a_{1} + 1) (a_{2} + 1) \dots (a_{k} + 1)$ .

1. What Is a Combination?

A combination is a selection of objects where the order does not matter. Choosing

{A, B, C}

is the same as choosing

{C, A, B}

; both count as one combination.

Contrast with permutation: ABC, ACB, BAC, BCA, CAB, CBA are 6 different permutations but only 1 combination. Formally, every combination of $r$ objects can be rearranged in $r!$ ways to produce permutations, so:

$^{n} P_{r} = r! \cdot^{n} C_{r} ⟺^{n} C_{r} = \frac{^{n} P _{r}}{r !} = \frac{n !}{r ! ( n - r )!}$

1.1 Selection Without Repetition

The number of ways to select

r

objects from

n

distinct objects, without repetition:

^{n} C_{r} = \frac{n !}{r ! ( n - r )!}, 0 \leq r \leq n

1.2 Selection With Repetition

The number of ways to select

r

objects from

n

distinct objects when each object may be chosen up to

r

times:

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

Solved Example 1

15 toys are to be distributed among 3 children. Any child may receive any number of toys. Find the number of ways if (i) the toys are distinct, (ii) the toys are identical.

Solution:

(i) Distinct toys. Each of the 15 toys can independently go to any of the 3 children. This is arrangement with repetition: $3^{15}$ ways.

(ii) Identical toys. The question reduces to selecting a child 15 times from 3 children (with repetition allowed): $^{3} H_{15} =^{3 + 15 - 1} C_{15} =^{17} C_{15} =^{17} C_{2} = 136$ ways.

2. Restricted Selection and Arrangement

Many problems fix conditions on specific objects: they must be included, or must be excluded. Handle these by splitting the selection.

Selecting

r

objects from

n

distinct objects when

k

particular objects are:

Always included: $^{n - k} C_{r - k}$ (the $k$ are fixed; choose the remaining $r - k$ from the other $n - k$ ).
Never included: $^{n - k} C_{r}$ (exclude the $k$ ; choose all $r$ from the other $n - k$ ).

For arrangements (order matters), multiply either count by

r!

Solved Example 2

A delegation of 4 students is to be selected from 12 students. In how many ways can it be done if:

All students are equally willing.
Two particular students must be included.
Two particular students refuse to be together.
Two particular students agree to serve only together (either both in or both out).
Students A and B want to serve together, while students C and D refuse to serve together.

Solution:

(a) $^{12} C_{4} = 495$ .

(b) The 2 particular students are fixed; choose 2 more from the other 10: $^{10} C_{2} = 45$ .

(c) Total minus (both selected) $= 495 - 45 = 450$ .

(d) Either both in ( $^{10} C_{2} = 45$ ) or both out ( $^{10} C_{4} = 210$ ). Total $= 45 + 210 = 255$ .

(e) Split on the status of ${A, B}$ (must both be in or both be out) and ${C, D}$ (not both in). The other 8 students are neutral. Six mutually exclusive cases:

${A, B, C}$ in, D out: pick 1 more from 8 $=^{8} C_{1} = 8$ .
${A, B, D}$ in, C out: pick 1 more from 8 $=^{8} C_{1} = 8$ .
${A, B}$ in, both C and D out: pick 2 more from 8 $=^{8} C_{2} = 28$ .
${A, B}$ out, C in, D out: pick 3 from 8 $=^{8} C_{3} = 56$ .
${A, B}$ out, D in, C out: pick 3 from 8 $=^{8} C_{3} = 56$ .
${A, B, C, D}$ all out: pick 4 from 8 $=^{8} C_{4} = 70$ .

Total $= 8 + 8 + 28 + 56 + 56 + 70 = 226$ .

3. Properties of $^{n} C_{r}$

Symmetry: $^{n} C_{r} =^{n} C_{n - r}$ . Choosing $r$ to include is the same as choosing $n - r$ to exclude.
Equality condition: If $^{n} C_{r} =^{n} C_{k}$ , then either $r = k$ or $r + k = n$ .
Pascal's identity: $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ . This is the building block of Pascal's triangle.
Recursion by $r$ : $^{n} C_{r} = \frac{n}{r} \cdot^{n - 1} C_{r - 1}$ .
Ratio of successive terms: $\frac{^{n} C _{r}}{^{n} C _{r - 1}} = \frac{n - r + 1}{r}$ .
Maximum value: $^{n} C_{r}$ is greatest at $r = n /2$ (if $n$ is even) or at $r = \frac{n - 1}{2}$ and $r = \frac{n + 1}{2}$ (if $n$ is odd).

4. Geometric Applications

Figure 1: For a hexagon (

n = 6

): total lines from pairs of vertices

=^{6} C_{2} = 15

; sides

= 6

; diagonals

= 15 - 6 = 9

; triangles

=^{6} C_{3} = 20

Solved Example 3

In an

n

-sided polygon (

n > 3

How many diagonals are there?
How many triangles can be formed by joining vertices? Of these, how many have (i) exactly one side common with the polygon, (ii) exactly two sides common, (iii) no sides common?

Solution:

(a) Lines joining pairs of vertices $=^{n} C_{2} = \frac{n ( n - 1 )}{2}$ . Subtract the $n$ sides to get diagonals: $\frac{n ( n - 1 )}{2} - n = \frac{n ( n - 3 )}{2}$ .

(b) Triangles from any 3 vertices $=^{n} C_{3} = \frac{n ( n - 1 ) ( n - 2 )}{6}$ .

(i) Exactly one side common: Fix a side (there are $n$ sides). The third vertex must not be adjacent to either endpoint of that side, so it is chosen from $n - 4$ vertices. Total $= n (n - 4)$ .

(ii) Exactly two sides common: The triangle must use three consecutive vertices $(A_{i}, A_{i + 1}, A_{i + 2})$ . There are $n$ such triples.

(iii) No sides common: Subtract (i) and (ii) from total: $^{n} C_{3} - n (n - 4) - n = \frac{n ( n - 1 ) ( n - 2 )}{6} - n (n - 4) - n = \frac{n ( n - 4 ) ( n - 5 )}{6}$

5. Division Into Groups

Unequal groups. Number of ways to divide

n

distinct objects into named groups of sizes

p, q, r

(with

p + q + r = n

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

Equal groups. If groups are of the same size and are unnamed (identical), divide further by the factorial of the number of equal-sized groups.

Solved Example 4

In how many ways can 12 distinct books be divided (a) among 3 students so that each gets 4 books, (b) into 3 equal bundles of 4 books each (bundles are unnamed)?

Solution:

(a) Students are distinguishable (named recipients). Number of ways $= \frac{12 !}{4 ! 4 ! 4 !} = 34650$ .

(b) The 3 bundles of equal size are indistinguishable, so we divide by $3!$ : $\frac{12 !}{4 ! 4 ! 4 ! \cdot 3 !} = \frac{34650}{6} = 5775$ .

6. Distribution of Identical Objects (Stars and Bars)

To distribute $n$ identical objects into $r$ distinct boxes, we count non-negative (or positive) integer solutions of $x_{1} + x_{2} + \dots + x_{r} = n$ .

Figure 2: Stars and bars. Arranging

n

identical stars and

(r - 1)

bars in a row corresponds one-to-one with distributions into

r

boxes.

Each box may be empty (non-negative solutions of $x_{1} + \dots + x_{r} = n$ ): $^{n + r - 1} C_{r - 1}$ .
Each box must contain at least one object (positive solutions): $^{n - 1} C_{r - 1}$ .

Solved Example 5

Find the number of ways to distribute 10 identical chocolates among 4 children so that (a) any child may get zero or more, (b) every child gets at least one.

Solution:

(a) Non-negative solutions of $x_{1} + x_{2} + x_{3} + x_{4} = 10$ : $^{10 + 4 - 1} C_{4 - 1} =^{13} C_{3} = 286$ .

(b) Positive solutions: $^{10 - 1} C_{4 - 1} =^{9} C_{3} = 84$ .

7. Number of Divisors

N = p_{1}^{a_{1}} p_{2}^{a_{2}} \dots p_{k}^{a_{k}}

is the prime factorisation of

N

, then the total number of positive divisors of

N

(a_{1} + 1) (a_{2} + 1) \dots (a_{k} + 1)

Each divisor selects an exponent for each prime independently, giving the product of choices.

Solved Example 6

Find the number of divisors of

360

Solution:

Prime factorisation: $360 = 2^{3} \times 3^{2} \times 5^{1}$ . Number of divisors $= (3 + 1) (2 + 1) (1 + 1) = 4 \times 3 \times 2 = 24$ .

8. Total Number of Selections

Selections from

n

distinct objects (taking at least one):

^{n} C_{1} +^{n} C_{2} +^{n} C_{3} + \dots +^{n} C_{n} = 2^{n} - 1

This follows from the binomial expansion

(1 + 1)^{n} = \sum_{r = 0}^{n}^{n} C_{r} = 2^{n}

; subtract the empty selection.

Solved Example 7

A person has 5 friends. In how many ways can he invite one or more of them to dinner?

Solution:

Each friend is independently either invited or not, giving $2^{5} = 32$ possibilities. Subtract the one case in which nobody is invited: $32 - 1 = 31$ ways.

8.1 Selection From Mixed (Identical + Distinct) Objects

Total selections (taking at least one) from

n

distinct objects and groups of identical objects (say

p

alike of one kind,

q

alike of another kind,

r

alike of a third kind):

(p + 1) (q + 1) (r + 1) 2^{n} - 1

For each identical group, choose how many to include (0 to

p

, giving

p + 1

options); for the

n

distinct objects, choose a subset (

2^{n}

ways). Subtract the empty selection.

Solved Example 8

A basket contains 3 identical red balls, 4 identical green balls, and 2 distinct yellow balls. In how many ways can one or more balls be selected?

Solution:

Red: $(3 + 1) = 4$ options. Green: $(4 + 1) = 5$ options. Yellow (distinct): $2^{2} = 4$ options. Total including the empty selection: $4 \times 5 \times 4 = 80$ . Subtract the empty case: $80 - 1 = 79$ ways.

Common Mistakes to Avoid

Watch out

Confusing selection with arrangement: If order matters, use $^{n} P_{r}$ ; if not, use $^{n} C_{r}$ . "Form a team of 4" is combination; "arrange 4 in a line" is permutation.
Missing the extra divide for equal groups: Dividing 12 books into 3 unnamed piles of 4 requires dividing by $3!$ ; dividing among 3 students does not.
Confusing "at least one" with total subsets: Total subsets of an $n$ -element set is $2^{n}$ ; non-empty selections is $2^{n} - 1$ .
Double counting in triangle problems: When counting triangles with restrictions, verify that (no side) + (one side) + (two sides) $=^{n} C_{3}$ .
Wrong stars-and-bars formula: "At least one" uses $^{n - 1} C_{r - 1}$ ; "may be zero" uses $^{n + r - 1} C_{r - 1}$ . Mixing these up is a common exam trap.
Assuming $^{n} C_{r} =^{n} C_{k}$ implies $r = k$ : The other solution $r + k = n$ is easy to overlook.

Frequently Asked Questions

Q1. What is the difference between permutation and combination?

A permutation is an ordered arrangement (order matters); a combination is an unordered selection (order does not matter). $^{n} P_{r}$ counts ordered arrangements, $^{n} C_{r}$ counts unordered selections, and $^{n} P_{r} = r! \cdot^{n} C_{r}$ .

Q2. When is $^{n} C_{r}$ used instead of $^{n} P_{r}$ ?

Use $^{n} C_{r}$ when the question asks to choose, select, or form a group/committee/team, and the order within the selection does not matter. Use $^{n} P_{r}$ when arranging, seating, ranking, or forming numbers or words.

Q3. Why is $^{n} C_{r} =^{n} C_{n - r}$ ?

Every time we choose $r$ objects to include, we equivalently choose $n - r$ objects to exclude. The two selections determine each other, so the counts are the same. This symmetry is often used to simplify calculations, e.g., $^{20} C_{17} =^{20} C_{3} = 1140$ .

Q4. What is Pascal's identity and why does it matter?

Pascal's identity states $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ . Combinatorially, when choosing $r$ from $n + 1$ objects, either the last object is included ( $^{n} C_{r - 1}$ ) or excluded ( $^{n} C_{r}$ ). This identity generates Pascal's triangle and appears in binomial theorem proofs.

Q5. How do I count triangles with vertices on a polygon?

Any 3 non-collinear vertices form a triangle, so the total is $^{n} C_{3}$ for an $n$ -sided polygon (all vertices are non-collinear on the polygon). Restrictions like "exactly one side common with the polygon" require careful subtraction: fix a side ( $n$ ways) and pick a third vertex not adjacent to it ( $n - 4$ ways), giving $n (n - 4)$ .

Q6. Why do we divide by $3!$ when splitting 12 books into 3 equal groups?

Because the 3 groups are of the same size and are unlabelled. Simply computing $\frac{12 !}{4 ! 4 ! 4 !}$ treats the 3 groups as if they were labelled (Group 1, Group 2, Group 3), overcounting by a factor of $3!$ - the number of ways to permute the labels. Dividing by $3!$ corrects this.

Q7. How do I find the number of divisors of a number?

Prime-factorise the number as $N = p_{1}^{a_{1}} p_{2}^{a_{2}} \dots p_{k}^{a_{k}}$ . The number of positive divisors is $(a_{1} + 1) (a_{2} + 1) \dots (a_{k} + 1)$ . Each divisor picks an exponent for each prime independently, from $0$ to $a_{i}$ .

Q8. What does "stars and bars" mean in JEE?

Stars and bars is a visual technique for distributing $n$ identical objects into $r$ distinct boxes. Represent objects as stars and box-dividers as bars. The number of arrangements of $n$ stars and $r - 1$ bars is $^{n + r - 1} C_{r - 1}$ , which equals the number of ways to distribute the objects.

Q9. Is combinations important for JEE Mains?

Yes. Combinations appear in 1-2 direct questions per year in JEE Mains and underpin Probability and Binomial Theorem questions. Common patterns include committee/team selection, geometry (diagonals, triangles), distribution problems, and identities like $^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$ .

Previous year questions on Combinations

37 questions from past papers, each with a step-by-step solution.

Show all 37 questions

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1. What Is a Combination?

1.1 Selection Without Repetition

1.2 Selection With Repetition

2. Restricted Selection and Arrangement

3. Properties of nCr​

4. Geometric Applications

5. Division Into Groups

6. Distribution of Identical Objects (Stars and Bars)

7. Number of Divisors

8. Total Number of Selections

8.1 Selection From Mixed (Identical + Distinct) Objects

Common Mistakes to Avoid

Frequently Asked Questions

Q1. What is the difference between permutation and combination?

Q2. When is nCr​ used instead of nPr​?

Q3. Why is nCr​=nCn−r​?

Q4. What is Pascal's identity and why does it matter?

Q5. How do I count triangles with vertices on a polygon?

Q6. Why do we divide by 3! when splitting 12 books into 3 equal groups?

Q7. How do I find the number of divisors of a number?

Q8. What does "stars and bars" mean in JEE?

Q9. Is combinations important for JEE Mains?

Previous year questions on Combinations

Ready to master Permutations And Combinations?

3. Properties of $^{n} C_{r}$

Q2. When is $^{n} C_{r}$ used instead of $^{n} P_{r}$ ?

Q3. Why is $^{n} C_{r} =^{n} C_{n - r}$ ?

Q6. Why do we divide by $3!$ when splitting 12 books into 3 equal groups?