Introduction to nCr (Combinations)
The nCr formula, also known as the combination formula, calculates the number of ways to choose r items from n items where order does not matter. This fundamental concept is essential in combinatorics, probability theory, and statistical analysis.
Core Formulas
| Formula Type | Mathematical Expression | Alternative Notation | Explanation |
|---|---|---|---|
| Basic nCr Formula | nCr = n! / (r! × (n-r)!) | C(n,r) = n! / (r! × (n-r)!) | Number of combinations of n items taken r at a time |
| nPr Formula | nPr = n! / (n-r)! | P(n,r) = n! / (n-r)! | Number of permutations of n items taken r at a time |
| Relationship Formula | nCr = nPr / r! | C(n,r) = P(n,r) / r! | Combinations equals permutations divided by r factorial |
| Factorial Definition | n! = n × (n-1) × (n-2) × … × 2 × 1 | 0! = 1 (by definition) | Product of all positive integers up to n |
Essential Properties and Special Cases
| Property/Case | Formula | Explanation | Example |
|---|---|---|---|
| Symmetry Property | nCr = nC(n-r) | Choosing r items equals choosing (n-r) items to exclude | 5C2 = 5C3 = 10 |
| Boundary Cases | nC0 = nCn = 1 | Only one way to choose nothing or everything | 7C0 = 7C7 = 1 |
| Single Selection | nC1 = n | n ways to choose one item from n items | 8C1 = 8 |
| Pascal’s Identity | nCr = (n-1)Cr + (n-1)C(r-1) | Used to construct Pascal’s triangle | 5C2 = 4C2 + 4C1 |
| Sum Property | Σ(nCr) = 2ⁿ (r=0 to n) | Sum of all combinations equals 2 to the power n | 3C0 + 3C1 + 3C2 + 3C3 = 8 = 2³ |
Probability Applications
| Application | Formula | Use Case | Example |
|---|---|---|---|
| Basic Probability | P(Event) = nCr / Total Combinations | Calculating probability using combinations | P(2 heads in 4 flips) = 4C2 / 2⁴ |
| Binomial Probability | P(X = k) = nCk × p^k × (1-p)^(n-k) | Probability of exactly k successes in n trials | P(exactly 3 successes in 10 trials) |
| Hypergeometric | P(X = k) = (KCk × (N-K)C(n-k)) / NCn | Sampling without replacement | Drawing cards from a deck |
Advanced Relationships
| Relationship | Formula | Mathematical Context |
|---|---|---|
| Binomial Theorem | (a + b)ⁿ = Σ(nCr × aⁿ⁻ʳ × bʳ) | Expansion of binomial expressions |
| Vandermonde’s Identity | (m+n)Cr = Σ(mCk × nC(r-k)) | Sum over all valid k values |
| Chu-Vandermonde | Σ(mCr × nCs) = (m+n)C(r+s) | When summing over specific ranges |
Computational Formulas for Large Numbers
| Method | Formula | Advantage |
|---|---|---|
| Multiplicative Form | nCr = (n × (n-1) × … × (n-r+1)) / (r × (r-1) × … × 1) | Avoids large factorial calculations |
| Recursive Formula | nCr = (n × (n-1)C(r-1)) / r | Efficient for sequential calculations |
| Logarithmic Form | log(nCr) = log(n!) – log(r!) – log((n-r)!) | For very large numbers to prevent overflow |
Common Mistakes
| Error Type | Incorrect | Correct | Note |
|---|---|---|---|
| Order Confusion | Using nPr for unordered selection | Use nCr for combinations | Order doesn’t matter in combinations |
| Zero Factorial | 0! = 0 | 0! = 1 | By mathematical convention |
| Negative Values | nCr where r > n | nCr = 0 when r > n | Cannot choose more than available |
| Non-integer Values | Using decimals for n or r | Only use non-negative integers | Combinations require whole numbers |
Quick Reference Examples
Basic Calculations
- 5C3 = 5!/(3! × 2!) = (5 × 4)/(2 × 1) = 10
- 7C2 = 7!/(2! × 5!) = (7 × 6)/(2 × 1) = 21
- 10C0 = 1 (choosing nothing)
- 10C10 = 1 (choosing everything)
Real-World Applications
- Team Selection: From 15 players, choose 11 for a team = 15C11 = 1,365
- Card Combinations: Choose 5 cards from 52 = 52C5 = 2,598,960
- Committee Formation: Choose 4 members from 20 people = 20C4 = 4,845
Memory Aids and Tips
- “Choose, don’t arrange” – Remember nCr is for selection without regard to order
- Symmetry shortcut – If r > n/2, calculate nC(n-r) instead for easier computation
- Pascal’s Triangle – Each entry is the sum of the two entries above it
- Multiplication principle – For sequential choices, multiply combinations
Differences: nCr vs nPr
| Aspect | nCr (Combinations) | nPr (Permutations) |
|---|---|---|
| Order | Order does NOT matter | Order DOES matter |
| Formula | n!/(r!(n-r)!) | n!/(n-r)! |
| Result | Always ≤ nPr | Always ≥ nCr |
| Example | Selecting team members | Arranging people in line |
Frequently Asked Questions (FAQs)
Q. What is the difference between nCr and nPr?
The fundamental difference lies in whether order matters:
- nCr (Combinations): Order does NOT matter. Used when selecting items where arrangement is irrelevant.
- Example: Choosing 3 students from 10 for a committee = 10C3 = 120
- nPr (Permutations): Order DOES matter. Used when arrangement is important.
- Example: Arranging 3 students in ranked positions = 10P3 = 720
Relationship: nPr = nCr × r! (permutations are always greater than or equal to combinations)
Q. How do I calculate nCr without a calculator?
Use the multiplicative method for efficient manual calculation:
Step-by-step approach:
- Write the multiplication form: nCr = (n × (n-1) × … × (n-r+1)) / (r × (r-1) × … × 1)
- Cancel common factors before multiplying
Example: Calculate 8C3
- Formula: (8 × 7 × 6) / (3 × 2 × 1)
- Cancel: (8 × 7 × 6) / 6 = 8 × 7 / 1 = 56
Q. Why is 0! (zero factorial) equal to 1?
0! = 1 by mathematical convention and logical consistency:
Logical reasons:
- Empty product definition: The product of no numbers equals 1 (multiplicative identity)
- Consistency with nCn: For nCn = n!/(n! × 0!) to equal 1, we need 0! = 1
- Recursive property: Since n! = n × (n-1)!, then 1! = 1 × 0!, so 0! must equal 1
Practical application: nC0 = n!/(0! × n!) = 1 (there’s exactly one way to choose nothing)
Q. When should I use nCr in real-life situations?
Use nCr whenever you need to select or choose items where order doesn’t matter:
Common applications:
- Lottery systems: Choosing 6 numbers from 49 = 49C6
- Team formation: Selecting 11 players from 15 for a football team = 15C11
- Menu combinations: Choosing 3 toppings from 10 available = 10C3
- Card games: Selecting 5 cards from a 52-card deck = 52C5
- Committee selection: Choosing 4 members from 20 people = 20C4
- Investment portfolios: Selecting 5 stocks from 50 options = 50C5
Key indicator: If rearranging your selection doesn’t create a new outcome, use nCr.
Q. What happens when r > n (r is greater than n)?
When r > n, nCr = 0 (mathematically undefined in standard contexts)
Logical explanation:
- You cannot choose more items than what’s available
- Example: You can’t select 5 apples from a basket of 3 apples
Mathematical handling:
- Standard definition: nCr = 0 when r > n
- The formula n!/(r!(n-r)!) becomes invalid because (n-r)! would be the factorial of a negative number
Important note: In standard combinatorics, both n and r must be non-negative integers with r ≤ n.
Q. How do I solve equations like nCr = nPr?
Answer: Use the relationship nCr = nPr / r! and solve algebraically:
Given: nCr = nPr
Solution process:
- Substitute formulas: n!/(r!(n-r)!) = n!/(n-r)!
- Simplify: 1/r! = 1
- Therefore: r! = 1
- Answer: r = 0 or r = 1
Verification:
- When r = 0: nC0 = nP0 = 1
- When r = 1: nC1 = nP1 = n
Insight: nCr equals nPr only when selecting 0 or 1 item, because arrangement doesn’t create new outcomes.
Q. Can nCr give a decimal or fractional answer?
No, nCr always gives a whole number (non-negative integer) when n and r are valid inputs.
Mathematical guarantee:
- Combinations count distinct selections, which must be whole numbers
- The formula n!/(r!(n-r)!) is designed to always yield an integer when 0 ≤ r ≤ n
Common misconception: Students sometimes get decimals due to:
- Calculation errors (incorrect order of operations)
- Using non-integer values for n or r (invalid inputs)
- Incorrect formula application
Validation check: If you get a decimal answer, recheck your calculation – you’ve made an error.
Q. What is Pascal’s Triangle and how does it relate to nCr?
Pascal’s Triangle is a triangular array where each number equals the sum of the two numbers above it, and each entry represents a combination value.
Structure and nCr relationship:
Row 0: 1 (0C0)
Row 1: 1 1 (1C0, 1C1)
Row 2: 1 2 1 (2C0, 2C1, 2C2)
Row 3: 1 3 3 1 (3C0, 3C1, 3C2, 3C3)
Row 4: 1 4 6 4 1 (4C0, 4C1, 4C2, 4C3, 4C4)
Main properties:
- Row n, Position r = nCr
- Pascal’s Identity: nCr = (n-1)Cr + (n-1)C(r-1)
- Binomial expansion: Row n gives coefficients of (a+b)ⁿ
- Quick calculation: Use the triangle to find combinations without direct calculation