Real-Life Hook: The Lottery Dilemma
You’re playing a lottery where you must choose 6 numbers from 1 to 49. In how many different ways can you select your numbers?
Wrong thinking: “I choose 1st number (49 options), 2nd number (48 options)… so $49 \times 48 \times 47 \times 46 \times 45 \times 44$”
Reality check: Does it matter if you pick {3,7,15,22,31,48} vs {48,31,22,15,7,3}? NO! Both are the same lottery ticket!
Correct answer:
$$^{49}C_6 = \frac{49!}{6! \times 43!} = 13,983,816 \text{ ways}$$Interactive Demo: Visualize Combinations
See how selections work when order doesn’t matter.
This is a combination problem - order doesn’t matter. Welcome to the world of selections!
Why This Topic Matters
Combinations appear everywhere:
- Team selection: Choosing 11 players from 15 for cricket
- Committee formation: Selecting 5 members from 20
- Lottery: Picking winning numbers
- Card games: Dealing hands of cards
- Subsets: Selecting items from a set
JEE Importance:
- JEE Main: 4-6 questions (highest weightage in P&C), 12-18 marks
- JEE Advanced: Complex selection with constraints, often with multinomial
- Foundation: Critical for probability, binomial theorem, distributions
The Big Difference: Permutation = Ordered, Combination = Unordered
Interactive Demo: Combination Calculator
Combination (nCr) Calculator
Calculate nCr = n!/(r!(n-r)!) - the number of ways to select r objects from n objects
Core Concepts
1. Definition of Combination
Combination: A selection of objects where order does not matter.
Notation: The number of ways to select $r$ objects from $n$ distinct objects is:
$$^nC_r \quad \text{or} \quad C(n,r) \quad \text{or} \quad \binom{n}{r} \quad \text{or} \quad C_r^n$$Read as: “n choose r”
2. Combination Formula
$$\boxed{^nC_r = \frac{n!}{r! \times (n-r)!}}$$Derivation:
- Number of ways to select and arrange $r$ from $n$: $^nP_r = \frac{n!}{(n-r)!}$
- But in combinations, order doesn’t matter
- Any $r$ objects can be arranged in $r!$ ways (all counting as the same selection)
- So divide by $r!$: $$^nC_r = \frac{^nP_r}{r!} = \frac{n!}{r! \times (n-r)!}$$
Example: Select 3 people from {A,B,C,D,E}
$$^5C_3 = \frac{5!}{3! \times 2!} = \frac{120}{6 \times 2} = 10$$The 10 selections: {A,B,C}, {A,B,D}, {A,B,E}, {A,C,D}, {A,C,E}, {A,D,E}, {B,C,D}, {B,C,E}, {B,D,E}, {C,D,E}
3. Important Properties of Combinations
Property 1: Symmetry
$$\boxed{^nC_r = ^nC_{n-r}}$$Proof:
$$^nC_r = \frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!r!} = ^nC_{n-r}$$Intuition: Choosing $r$ objects to include = Choosing $(n-r)$ objects to exclude
Example: $^{10}C_3 = ^{10}C_7 = 120$
Property 2: Sum of consecutive combinations
$$\boxed{^nC_r + ^nC_{r-1} = ^{n+1}C_r}$$Proof: (Pascal’s triangle property)
$$^nC_r + ^nC_{r-1} = \frac{n!}{r!(n-r)!} + \frac{n!}{(r-1)!(n-r+1)!}$$After simplification (try it!), this equals $^{n+1}C_r$.
Intuition: Divide selections from $n+1$ objects into two groups:
- Selections that include a specific object: $^nC_{r-1}$
- Selections that exclude that object: $^nC_r$
Property 3: Sum of all combinations
$$\boxed{^nC_0 + ^nC_1 + ^nC_2 + \ldots + ^nC_n = 2^n}$$Proof: From binomial theorem, $(1+1)^n = 2^n$
Intuition: Total number of subsets of a set with $n$ elements = $2^n$
Property 4: Special values
$$\boxed{^nC_0 = 1 \quad ^nC_1 = n \quad ^nC_n = 1}$$Property 5: Maximum value
- If $n$ is even: $^nC_{n/2}$ is maximum
- If $n$ is odd: $^nC_{(n-1)/2} = ^nC_{(n+1)/2}$ are maximum
Example: For $n=6$: $^6C_3 = 20$ is maximum For $n=7$: $^7C_3 = ^7C_4 = 35$ are maximum
4. Relation Between nPr and nCr
$$\boxed{^nP_r = ^nC_r \times r!}$$Interpretation:
- $^nC_r$ = Ways to select $r$ objects (no order)
- $r!$ = Ways to arrange those $r$ objects
- $^nP_r$ = Select AND arrange = $^nC_r \times r!$
Example:
$$^5P_3 = ^5C_3 \times 3! = 10 \times 6 = 60$$Memory Tricks
nCr vs nPr - Decision Tree
START
↓
Does ORDER matter?
↙ ↘
YES NO
↓ ↓
nPr nCr
(Arrange) (Select)
Quick questions to ask:
- “If I swap two items, is it different?” → YES = Permutation, NO = Combination
- “Am I creating a ranking/sequence?” → YES = Permutation
- “Am I forming a team/group/committee?” → YES = Combination
Memory Hooks
“Committee = Combination” (both start with C) “Selection = Set = Combination” (sets are unordered)
“Arrangement = Order = Permutation” “Race = Rank = Permutation”
The Formula Memory Trick
$$^nC_r = \frac{n!}{r! \times (n-r)!}$$Mnemonic: “Cut the factorial twice - by $r!$ and $(n-r)!$”
Compared to $^nP_r = \frac{n!}{(n-r)!}$ which cuts once.
nCr has EXTRA division by $r!$ → Removes order
Pascal’s Triangle (Visual Memory)
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
- Row $n$ contains $^nC_0, ^nC_1, \ldots, ^nC_n$
- Each number = sum of two numbers above it
- Symmetric (property: $^nC_r = ^nC_{n-r}$)
Common Counting Mistakes
❌ Mistake 1: Using nPr Instead of nCr
Problem: Select 3 students from 10 to form a study group.
Wrong: $^{10}P_3 = 720$ ❌ (This counts ABC different from BAC!)
Right: $^{10}C_3 = 120$ ✓ (Group has no order)
Test: Ask yourself - is {Alice, Bob, Charlie} different from {Charlie, Bob, Alice}? NO → Use nCr!
❌ Mistake 2: Not Simplifying nCr
Problem: Calculate $^{50}C_{48}$
Wrong: Try to compute $\frac{50!}{48! \times 2!}$ directly ❌
Right: Use symmetry! $^{50}C_{48} = ^{50}C_2 = \frac{50 \times 49}{2} = 1,225$ ✓
Rule: If $r > n/2$, use $^nC_r = ^nC_{n-r}$ for easier calculation
❌ Mistake 3: Forgetting “At Least One”
Problem: Select at least 2 people from 5.
Wrong: $^5C_2 = 10$ ❌ (This is EXACTLY 2, not at least!)
Right: $^5C_2 + ^5C_3 + ^5C_4 + ^5C_5 = 10 + 10 + 5 + 1 = 26$ ✓
Alternative: Total - Unwanted = $(2^5 - 1) - ^5C_0 - ^5C_1 = 31 - 1 - 5 = 25$
Wait, let me recalculate:
- Total subsets: $2^5 = 32$
- Subsets with 0 people: $^5C_0 = 1$
- Subsets with 1 person: $^5C_1 = 5$
- At least 2: $32 - 1 - 5 = 26$ ✓
❌ Mistake 4: Confusing “And” with “Or” in Selections
Problem: Select 2 boys AND 3 girls from 5 boys and 6 girls.
Wrong: $^5C_2 + ^6C_3 = 10 + 20 = 30$ ❌ (This is OR, not AND!)
Right: $^5C_2 \times ^6C_3 = 10 \times 20 = 200$ ✓ (AND means multiply!)
Problem-Solving Strategies
Strategy 1: Identify Selection vs Arrangement
Ask: Does order matter?
- No → Use $^nC_r$
- Yes → Use $^nP_r$
Strategy 2: Use Symmetry Property
For large $r$ close to $n$:
$$^nC_r = ^nC_{n-r}$$Example: $^{100}C_{98} = ^{100}C_2 = 4,950$ (much easier!)
Strategy 3: Complementary Counting
“At least”, “at most” → Use Total - Unwanted
Example: At least 1 person from 6
- Total ways: $2^6 - 1 = 63$ (all subsets except empty)
- OR: $^6C_1 + ^6C_2 + \ldots + ^6C_6 = 2^6 - ^6C_0 = 63$
Strategy 4: Break Into Cases
When selections have constraints, divide into non-overlapping cases.
Example: Select 4 from 6 men and 5 women, with at least 1 woman
- Case 1: 1W, 3M → $^5C_1 \times ^6C_3 = 5 \times 20 = 100$
- Case 2: 2W, 2M → $^5C_2 \times ^6C_2 = 10 \times 15 = 150$
- Case 3: 3W, 1M → $^5C_3 \times ^6C_1 = 10 \times 6 = 60$
- Case 4: 4W, 0M → $^5C_4 \times ^6C_0 = 5 \times 1 = 5$
- Total: $100 + 150 + 60 + 5 = 315$
Alternative: Total - (All men) = $^{11}C_4 - ^6C_4 = 330 - 15 = 315$ ✓
Practice Problems
Level 1: Foundation (JEE Main)
Problem 1.1: Find the value of $^8C_3$.
Solution
Answer: 56
Solution:
$$^8C_3 = \frac{8!}{3! \times 5!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$$Problem 1.2: In how many ways can a committee of 5 be selected from 8 people?
Solution
Answer: 56
Solution:
$$^8C_5 = \frac{8!}{5! \times 3!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56$$Note: $^8C_5 = ^8C_3 = 56$ (symmetry property!)
Problem 1.3: Find $n$ if $^nC_2 = 28$.
Solution
Answer: $n = 8$
Solution:
$$^nC_2 = \frac{n!}{2!(n-2)!} = \frac{n(n-1)}{2} = 28$$ $$n(n-1) = 56$$ $$n^2 - n - 56 = 0$$ $$(n-8)(n+7) = 0$$ $$n = 8 \text{ or } n = -7$$Since $n$ must be positive, $n = 8$.
Verification: $^8C_2 = \frac{8 \times 7}{2} = 28$ ✓
Problem 1.4: Prove that $^nC_r = ^nC_{n-r}$.
Solution
Proof:
$$^nC_r = \frac{n!}{r!(n-r)!}$$ $$^nC_{n-r} = \frac{n!}{(n-r)!(n-(n-r))!} = \frac{n!}{(n-r)!r!}$$Since multiplication is commutative:
$$\frac{n!}{r!(n-r)!} = \frac{n!}{(n-r)!r!}$$Therefore, $^nC_r = ^nC_{n-r}$ ✓
Intuitive proof: Selecting $r$ objects to include is the same as selecting $(n-r)$ objects to exclude.
Level 2: Intermediate (JEE Main/Advanced)
Problem 2.1: From 6 boys and 4 girls, a committee of 5 is to be formed. In how many ways can this be done if the committee contains: (a) Exactly 2 girls? (b) At least 2 girls? (c) At most 2 girls?
Solution
(a) Exactly 2 girls:
- Select 2 girls from 4: $^4C_2 = 6$
- Select 3 boys from 6: $^6C_3 = 20$
- Total: $6 \times 20 = 120$
(b) At least 2 girls: Cases: 2G+3B, 3G+2B, 4G+1B
- 2G, 3B: $^4C_2 \times ^6C_3 = 6 \times 20 = 120$
- 3G, 2B: $^4C_3 \times ^6C_2 = 4 \times 15 = 60$
- 4G, 1B: $^4C_4 \times ^6C_1 = 1 \times 6 = 6$
- Total: $120 + 60 + 6 = 186$
(c) At most 2 girls: Cases: 0G+5B, 1G+4B, 2G+3B
- 0G, 5B: $^4C_0 \times ^6C_5 = 1 \times 6 = 6$
- 1G, 4B: $^4C_1 \times ^6C_4 = 4 \times 15 = 60$
- 2G, 3B: $^4C_2 \times ^6C_3 = 6 \times 20 = 120$
- Total: $6 + 60 + 120 = 186$
Alternative for (c): Total - (At least 3 girls) = $^{10}C_5 - (^4C_3 \times ^6C_2 + ^4C_4 \times ^6C_1)$ = $252 - (60 + 6) = 186$ ✓
Problem 2.2: Out of 7 consonants and 4 vowels, how many words of 3 consonants and 2 vowels can be formed?
Solution
Answer: 25,200
Solution:
Step 1: Select 3 consonants from 7
$$^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$Step 2: Select 2 vowels from 4
$$^4C_2 = \frac{4 \times 3}{2} = 6$$Step 3: Arrange these 5 letters (3 consonants + 2 vowels)
$$5! = 120$$Total: $35 \times 6 \times 120 = 25,200$
Key insight: We SELECT letters (combination), then ARRANGE them (permutation).
Problem 2.3: In how many ways can a student select 5 courses out of 9 if 2 specific courses are compulsory?
Solution
Answer: 35
Solution:
Since 2 courses are compulsory, they must be included.
Remaining slots: $5 - 2 = 3$ Remaining courses: $9 - 2 = 7$
Select 3 from 7:
$$^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35$$Problem 2.4: Prove that $^nC_0 + ^nC_1 + ^nC_2 + \ldots + ^nC_n = 2^n$.
Solution
Proof (Binomial Theorem):
From binomial theorem:
$$(x + y)^n = \sum_{r=0}^{n} ^nC_r \cdot x^{n-r} \cdot y^r$$Put $x = 1, y = 1$:
$$(1 + 1)^n = \sum_{r=0}^{n} ^nC_r \cdot 1^{n-r} \cdot 1^r$$ $$2^n = \sum_{r=0}^{n} ^nC_r$$ $$2^n = ^nC_0 + ^nC_1 + ^nC_2 + \ldots + ^nC_n$$Proof (Combinatorial):
Consider a set with $n$ elements. Each element can either be included or excluded from a subset (2 choices per element).
Total subsets: $2^n$
Alternatively, count by size:
- Subsets of size 0: $^nC_0$
- Subsets of size 1: $^nC_1$
- …
- Subsets of size $n$: $^nC_n$
Total: $^nC_0 + ^nC_1 + \ldots + ^nC_n = 2^n$ ✓
Level 3: Advanced (JEE Advanced)
Problem 3.1: Find the number of ways to select 4 cards from a deck of 52 cards such that: (a) All are of the same suit (b) All are of different suits (c) Exactly 2 are of the same suit
Solution
(a) All of same suit:
- Choose 1 suit from 4: $^4C_1 = 4$
- Choose 4 cards from 13 of that suit: $^{13}C_4 = 715$
- Total: $4 \times 715 = 2,860$
(b) All of different suits:
- Each of the 4 cards from a different suit
- Choose 1 card from hearts: $^{13}C_1 = 13$
- Choose 1 card from diamonds: $^{13}C_1 = 13$
- Choose 1 card from clubs: $^{13}C_1 = 13$
- Choose 1 card from spades: $^{13}C_1 = 13$
- Total: $13^4 = 28,561$
(c) Exactly 2 of same suit: This is tricky! “Exactly 2 of same suit” is ambiguous. Let me interpret as “exactly one pair (2 cards of one suit, 2 other cards of different suits)”.
- Choose suit for the pair: $^4C_1 = 4$
- Choose 2 cards from that suit: $^{13}C_2 = 78$
- Choose 2 different suits for remaining 2 cards: $^3C_2 = 3$
- Choose 1 card from first suit: $^{13}C_1 = 13$
- Choose 1 card from second suit: $^{13}C_1 = 13$
- Total: $4 \times 78 \times 3 \times 13 \times 13 = 158,184$
Alternative interpretation: “At least one pair” This would require counting all configurations with pairs, which is complex.
For standard interpretation (exactly one pair of same suit, others different): Answer: 158,184
Problem 3.2: In how many ways can 11 books be divided between two students such that each gets at least 1 book?
Solution
Answer: 2,046
Solution:
Method 1 (Complementary counting):
- Total ways to distribute 11 books to 2 students (each book can go to either student): $2^{11} = 2,048$
- Subtract cases where one student gets all books: 2 (all to student 1 OR all to student 2)
- Answer: $2,048 - 2 = 2,046$
Method 2 (Case by case): Student 1 can get 1, 2, 3, …, or 10 books (not 0, not 11)
- Student 1 gets 1: $^{11}C_1 = 11$
- Student 1 gets 2: $^{11}C_2 = 55$
- …
- Student 1 gets 10: $^{11}C_{10} = 11$
Total: $^{11}C_1 + ^{11}C_2 + \ldots + ^{11}C_{10} = 2^{11} - ^{11}C_0 - ^{11}C_{11} = 2,048 - 2 = 2,046$
Note: If students are identical (indistinguishable), we’d divide by 2, but typically “between two students” means they’re distinguishable.
Answer: 2,046
Problem 3.3: Find the number of diagonals of a polygon with $n$ sides.
Solution
Answer: $\frac{n(n-3)}{2}$
Derivation:
A diagonal connects two non-adjacent vertices.
Total line segments joining any 2 vertices:
$$^nC_2 = \frac{n(n-1)}{2}$$Sides of polygon: $n$
Diagonals: Total line segments - Sides
$$= \frac{n(n-1)}{2} - n = \frac{n(n-1) - 2n}{2} = \frac{n(n-1-2)}{2} = \frac{n(n-3)}{2}$$Example: Hexagon ($n=6$)
$$\text{Diagonals} = \frac{6 \times 3}{2} = 9$$Verification: From each vertex, you can draw $(n-3)$ diagonals (can’t connect to itself or 2 adjacent vertices). Total: $n(n-3)$, but this counts each diagonal twice, so divide by 2.
$$\boxed{\frac{n(n-3)}{2}}$$Problem 3.4: If $^nC_r : ^nC_{r+1} = 1:2$ and $^nC_{r+1} : ^nC_{r+2} = 2:3$, find $n$ and $r$.
Solution
Answer: $n = 14, r = 4$
Solution:
From first condition:
$$\frac{^nC_r}{^nC_{r+1}} = \frac{1}{2}$$Using the formula:
$$\frac{^nC_r}{^nC_{r+1}} = \frac{n!/(r!(n-r)!)}{n!/((r+1)!(n-r-1)!)} = \frac{(r+1)!(n-r-1)!}{r!(n-r)!}$$ $$= \frac{r+1}{n-r} = \frac{1}{2}$$ $$2(r+1) = n-r$$ $$2r + 2 = n - r$$ $$3r + 2 = n \quad \ldots (1)$$From second condition:
$$\frac{^nC_{r+1}}{^nC_{r+2}} = \frac{2}{3}$$Similarly:
$$\frac{r+2}{n-r-1} = \frac{2}{3}$$ $$3(r+2) = 2(n-r-1)$$ $$3r + 6 = 2n - 2r - 2$$ $$5r + 8 = 2n \quad \ldots (2)$$Solve equations (1) and (2):
From (1): $n = 3r + 2$
Substitute in (2):
$$5r + 8 = 2(3r + 2)$$ $$5r + 8 = 6r + 4$$ $$4 = r$$So $r = 4$
From (1): $n = 3(4) + 2 = 14$
Answer: $n = 14, r = 4$
Verification:
- $^{14}C_4 = 1,001$
- $^{14}C_5 = 2,002$
- $^{14}C_6 = 3,003$
Check: $\frac{1001}{2002} = \frac{1}{2}$ ✓ Check: $\frac{2002}{3003} = \frac{2}{3}$ ✓
Cross-Topic Connections
Link to Binomial Theorem
The coefficients in the expansion of $(x+y)^n$ are:
$$(x+y)^n = ^nC_0 x^n + ^nC_1 x^{n-1}y + ^nC_2 x^{n-2}y^2 + \ldots + ^nC_n y^n$$Combinations are binomial coefficients!
→ See Binomial Theorem
Link to Probability
Many probability problems use combinations:
$$P(\text{Event}) = \frac{\text{Favorable selections}}{\text{Total selections}}$$Example: Probability of getting 2 aces in a 5-card hand:
$$P = \frac{^4C_2 \times ^{48}C_3}{^{52}C_5}$$→ See Probability
Link to Permutations
Conversion formula:
$$^nP_r = ^nC_r \times r!$$Selection → Combination Arrangement → Permutation Selection + Arrangement → $^nC_r \times r! = ^nP_r$
→ See Permutations Basics
Link to Set Theory
$^nC_r$ counts the number of subsets of size $r$ from a set of size $n$.
Total subsets: $\sum_{r=0}^{n} ^nC_r = 2^n$ (power set)
→ See Sets, Relations & Functions
JEE Tips & Tricks
Recognition Checklist
Is it a combination problem?
- Keywords: “select”, “choose”, “committee”, “team”, “group”
- “In how many ways” + no mention of order/arrangement
- “How many subsets/combinations”
- Does swapping items create the same result?
If ALL YES → Use $^nC_r$
Time-Saving Strategies
- Use symmetry: If $r > n/2$, compute $^nC_{n-r}$ instead
- Simplify before calculating: Cancel factorials, don’t expand!
- For “at least/most”: Use complementary counting
- For multiple constraints: Break into cases, then add
- Memorize small values: $^nC_2 = \frac{n(n-1)}{2}$, $^nC_3 = \frac{n(n-1)(n-2)}{6}$
Common JEE Patterns
- Committee/Team selection: With constraints (gender, subject, etc.)
- Card problems: Selecting from deck with suit/rank constraints
- Distribution: Dividing objects among people
- Geometry: Triangles from points, diagonals of polygons
- “At least/at most”: Use total - unwanted
Quick Mental Calculations
Memorize:
- $^nC_1 = n$
- $^nC_2 = \frac{n(n-1)}{2}$
- $^nC_{n-1} = n$
- $^nC_n = 1$
For small $n$:
- $^4C_2 = 6$
- $^5C_2 = 10$
- $^6C_2 = 15$
- $^6C_3 = 20$
Summary
| Aspect | Permutation | Combination |
|---|---|---|
| Meaning | Arrangement (ordered) | Selection (unordered) |
| Formula | $\frac{n!}{(n-r)!}$ | $\frac{n!}{r!(n-r)!}$ |
| Keywords | Arrange, order, rank | Select, choose, group |
| Example | Podium (1st, 2nd, 3rd) | Committee members |
| ABC vs BAC | Different | Same |
Key Formulas:
$$\boxed{^nC_r = \frac{n!}{r!(n-r)!}} \quad \boxed{^nC_r = ^nC_{n-r}} \quad \boxed{^nP_r = ^nC_r \times r!}$$Key Insight: Combinations remove order. When order doesn’t matter, divide by $r!$!
Next Steps:
- Master Combinations Applications for distribution and selection with constraints
- Learn Derangements for advanced permutation problems
- Apply to Probability and Binomial Theorem
Last updated: September 18, 2025