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Binomial Coefficient Calculator

Calculate C(n,k) = n!/(k!(n-k)!)

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📐Binomial Coefficient Formulas

Basic Formula
C(n,k) = n! / (k! × (n-k)!)
Combination formula
Alternative Notation
(n choose k) = nCk
Common notations
Symmetry Property
C(n,k) = C(n, n-k)
Useful for optimization
Pascal's Identity
C(n,k) = C(n-1,k-1) + C(n-1,k)
Recursive relation

💼Applications

Combinatorics
• Choosing combinations
• Counting problems
• Pascal's triangle
Probability
• Binomial distribution
• Probability calculations
• Statistical analysis
Algebra
• Binomial theorem
• Polynomial expansion
• Series calculations

💡Special Values

C(n, 0) = 1
C(n, 1) = n
C(n, n) = 1
C(n, 2) = n(n-1)/2
C(n, n-1) = n
C(n, k) = 0 if k > n

Binomial Coefficient Calculator: Calculate "N Choose K" Combinations

Table of Contents - Binomial Coefficients

How to Use This Calculator - Binomial Coefficients

Enter two values:

  • n (total number of items) - must be a non-negative integer
  • k (number of items to choose) - must be non-negative and not greater than n

Click "Calculate" to see:

  • The binomial coefficient C(n,k) or "n choose k"
  • The calculation using factorials
  • Step-by-step computation
  • Connection to Pascal's triangle
  • Practical interpretation

The calculator handles large values efficiently, avoiding factorial overflow for practical ranges.

Understanding Binomial Coefficients

Binomial coefficients answer a simple question: "In how many ways can I choose k items from n items when order doesn't matter?"

The everyday scenario:

You have 5 friends and want to invite 3 to dinner. How many different groups of 3 could you invite? That's C(5,3), read as "5 choose 3." The answer is 10—there are 10 different possible groups.

Why "binomial"?

They're called binomial coefficients because they appear as coefficients in the binomial theorem. When you expand (a + b)^n, the coefficients are binomial coefficients. For example:

(a + b)³ = 1a³ + 3a²b + 3ab² + 1b³

The coefficients 1, 3, 3, 1 are C(3,0), C(3,1), C(3,2), C(3,3).

The notation:

C(n,k) = "n choose k" Also written as ⁿCₖ or (n k) with parentheses arranged vertically

The formula:

C(n,k) = n! / (k! × (n-k)!)

Where n! (n factorial) means n × (n-1) × (n-2) × ... × 2 × 1.

Why this formula works:

Think about arranging k items from n items:

  • There are n choices for the first item
  • n-1 choices for the second
  • Continue until you've made k choices
  • This gives n!/(n-k)! arrangements

But we overcounted! If order doesn't matter, the k items can be arranged in k! ways, and all those arrangements represent the same combination. So we divide by k! to correct for overcounting.

Result: n! / (k! × (n-k)!)

Pascal's triangle connection:

Binomial coefficients form Pascal's triangle:

         1
       1   1
     1   2   1
   1   3   3   1
 1   4   6   4   1

Each number is the sum of the two numbers above it. The nth row contains the binomial coefficients for n.

Key properties:

Symmetry: C(n,k) = C(n, n-k) Choosing k items is the same as choosing which n-k items to leave out.

Sum: C(n,0) + C(n,1) + ... + C(n,n) = 2^n The total number of all possible subsets of n items is 2^n.

How to Calculate Binomial Coefficients Manually

Method 1: Direct factorial calculation (small values)

Example 1: C(5,2)

Formula: C(5,2) = 5! / (2! × 3!)

Step 1: Calculate 5! = 5 × 4 × 3 × 2 × 1 = 120

Step 2: Calculate 2! = 2 × 1 = 2

Step 3: Calculate 3! = 3 × 2 × 1 = 6

Step 4: C(5,2) = 120 / (2 × 6) = 120 / 12 = 10

Method 2: Cancellation method (more efficient)

Instead of computing large factorials, cancel common terms.

Example 2: C(10,3)

C(10,3) = 10! / (3! × 7!)

Step 1: Write out partially = (10 × 9 × 8 × 7!) / (3! × 7!)

Step 2: Cancel 7! = (10 × 9 × 8) / (3 × 2 × 1)

Step 3: Calculate numerator = 720 / 6

Step 4: Divide = 120

Example 3: C(52,5) - Poker hands

C(52,5) = 52! / (5! × 47!)

Cancel 47!: = (52 × 51 × 50 × 49 × 48) / (5 × 4 × 3 × 2 × 1)

Numerator = 311,875,200 Denominator = 120 Result = 2,598,960

(This is why there are about 2.6 million possible 5-card poker hands!)

Method 3: Pascal's triangle (for small values)

Build Pascal's triangle row by row. Each entry is the sum of the two entries above it.

To find C(4,2):

Row 0:     1
Row 1:    1 1
Row 2:   1 2 1
Row 3:  1 3 3 1
Row 4: 1 4 6 4 1

C(4,2) is the 3rd entry (counting from 1) in row 4, which is 6.

Method 4: Recursive formula

C(n,k) = C(n-1,k-1) + C(n-1,k)

This says: to choose k items from n items, either include the first item and choose k-1 from the remaining n-1, or don't include the first item and choose k from the remaining n-1.

Base cases: C(n,0) = 1 (one way to choose nothing) C(n,n) = 1 (one way to choose everything) C(n,k) = 0 if k greater than n

Example 4: Using symmetry

C(100,98) looks hard to compute, but use symmetry:

C(100,98) = C(100,2) = (100 × 99) / 2 = 4,950

Much easier than computing 98!

Real-World Applications

Lottery odds: A lottery requires choosing 6 numbers from 49. Total possible combinations = C(49,6) = 13,983,816. Your odds of winning are 1 in about 14 million.

Committee selection: A company has 12 employees and needs to form a 4-person committee. Number of possible committees = C(12,4) = 495.

Genetics: When parents each contribute chromosomes, the number of genetically unique children possible is astronomical. With 23 chromosome pairs, there are C(46,23) ways chromosomes can assort (over 8 billion).

Password security: If a password requires exactly 3 uppercase letters from 26 options in any order, there are C(26,3) = 2,600 ways to choose which letters, then multiply by arrangements.

Quality control: Testing 5 items from a batch of 100 for quality control. Number of different samples = C(100,5) = 75,287,520.

Tournament brackets: In a tournament with n teams where k advance to playoffs, there are C(n,k) possible playoff scenarios (before considering matchups).

Card games: In bridge, each player gets 13 cards from 52. Number of possible hands for one player = C(52,13) ≈ 635 billion.

Network connections: In a network of n devices where each device connects to k others, the number of possible connection patterns involves binomial coefficients.

Combinatorial optimization: Problems like "select k items to maximize value subject to weight constraint" explore C(n,k) possible selections.

Common Mistakes and How to Avoid Them

Mistake 1: Confusing permutations and combinations

Wrong: Thinking C(5,2) = 5 × 4 = 20

Right: C(5,2) = (5 × 4) / 2 = 10

Why: Permutations care about order (AB different from BA). Combinations don't (AB same as BA). Divide by k! to correct for overcounting.

Mistake 2: Computing huge factorials unnecessarily

Wrong: Computing C(100,2) = 100! / (2! × 98!) by calculating 100! = giant number

Right: C(100,2) = (100 × 99) / 2 = 4,950 (cancel common terms first)

Why: Factorial values explode quickly. Always cancel before computing.

Mistake 3: Using the formula when k greater than n

Wrong: Trying to compute C(5,7)

Right: C(5,7) = 0 (impossible to choose 7 items from only 5)

Why: You can't choose more items than you have.

Mistake 4: Forgetting k = 0 or k = n cases

Wrong: Thinking C(n,0) = 0

Right: C(n,0) = 1 (one way to choose nothing: choose nothing!)

Similarly: C(n,n) = 1 (one way to choose everything)

Mistake 5: Integer division errors in programming

Wrong: Computing n/k in integer division before multiplying

Right: Multiply first, then divide, or use floating point carefully

Why: Integer division truncates, losing precision.

Mistake 6: Not using symmetry

Wrong: Computing C(1000,997) using the formula as written

Right: C(1000,997) = C(1000,3) = much easier to compute

Why: Symmetry saves massive computation time.

Recovery strategy:

For small values, verify using Pascal's triangle. For large values, use cancellation and symmetry. Always check if your answer makes intuitive sense (e.g., C(5,2) should be smallish, not millions).

Related Topics

Permutations: When order matters, use P(n,k) = n!/(n-k)!. Related to combinations by P(n,k) = k! × C(n,k).

Binomial Theorem: (a+b)^n = Σ C(n,k) a^(n-k) b^k. Binomial coefficients are the coefficients in the expansion.

Pascal's Triangle: A triangular array where each number is the sum of the two above it. Contains all binomial coefficients.

Multinomial Coefficients: Generalization for choosing multiple groups instead of just one group.

Probability: Combinations are fundamental for calculating probabilities in situations with multiple outcomes.

How This Calculator Works

The calculator uses efficient algorithms to avoid overflow:

For small values:

Direct computation with factorial formula
C(n,k) = n! / (k! × (n-k)!)

For moderate values:

Cancellation method:
numerator = n × (n-1) × ... × (n-k+1)
denominator = k × (k-1) × ... × 1
result = numerator / denominator
(Compute by alternating multiply and divide to keep values manageable)

Using symmetry:

if k > n/2:
    C(n,k) = C(n, n-k)
(Reduces computation for large k)

For very large values:

Use logarithms to avoid overflow:
log(C(n,k)) = log(n!) - log(k!) - log((n-k)!)
Then exponentiate result

Optimization:

Base cases:
if k = 0 or k = n: return 1
if k > n: return 0

Use recursive formula for moderate values:
C(n,k) = C(n-1,k-1) + C(n-1,k)
with memoization to cache results

All calculations done locally in your browser.

FAQs

What does "n choose k" mean?

It's the number of ways to choose k items from n items when order doesn't matter. Like picking 3 people from 10 to form a team.

What's the difference between combinations and permutations?

Combinations don't care about order (AB same as BA). Permutations do (AB different from BA). C(n,k) = P(n,k) / k!.

Why is C(n,0) equal to 1?

There's exactly one way to choose nothing: don't choose anything! It's a valid selection of 0 items.

Can I have C(5,7)?

No, that equals 0. You can't choose 7 items from only 5. k must be less than or equal to n.

What is C(n,n)?

It equals 1. There's only one way to choose all n items: take all of them.

How do I compute C(100,50) without a calculator?

You don't—it's enormous (about 10^29). Use software or a calculator with big integer support.

What's the largest value in row n of Pascal's triangle?

C(n, n/2) if n is even, or C(n, (n±1)/2) if n is odd. The middle value(s) are largest due to symmetry.

Why do binomial coefficients appear in probability?

When calculating probabilities of k successes in n independent trials, you need to count the number of ways to achieve k successes, which is C(n,k).

How does C(n,k) relate to the binomial theorem?

In (a+b)^n = Σ C(n,k) a^(n-k) b^k, the C(n,k) are the coefficients. They tell you how many ways each term can occur.

What's the sum of all binomial coefficients in row n?

C(n,0) + C(n,1) + ... + C(n,n) = 2^n. This represents the total number of subsets of an n-element set.

Can n or k be negative?

In basic combinatorics, no. But generalized binomial coefficients allow negative or non-integer values using the gamma function.

What's the alternating sum of binomial coefficients?

C(n,0) - C(n,1) + C(n,2) - ... = 0 for n greater than 0. This comes from setting a=1, b=-1 in (a+b)^n.

How do I verify my answer?

Use symmetry: C(n,k) should equal C(n,n-k). Also check with Pascal's triangle for small values, or verify using the recursive formula.

What's the connection to subsets?

C(n,k) counts the number of k-element subsets of an n-element set. The power set (all subsets) has size 2^n.

Why does cancellation work?

n! / (n-k)! = n × (n-1) × ... × (n-k+1), because (n-k)! cancels the lower portion of n!. Then divide by k! to get C(n,k).

How are binomial coefficients used in statistics?

Binomial distribution, hypothesis testing, and experimental design all use C(n,k) to count outcomes.

What happens if I choose k = 1?

C(n,1) = n. There are n ways to choose 1 item from n items—just pick which one!

Can C(n,k) ever be a fraction?

No, for non-negative integer n and k. It always results in a whole number (a count of combinations).

What's the fastest way to compute C(n,k)?

Use symmetry if k greater than n/2, then use the cancellation method. For very large values, use software with optimized algorithms.

How do binomial coefficients grow?

They increase as you move toward the middle of a row in Pascal's triangle, then decrease symmetrically. The middle value(s) are largest.

Additional Notes

Binomial coefficients are fundamental in combinatorics, probability, and algebra. They bridge counting problems with algebraic expansions, revealing deep connections throughout mathematics.

Understanding when to use them (any "choose k from n" scenario) is more important than memorizing formulas. The key insight is recognizing that order doesn't matter—that's what distinguishes combinations from permutations.

Practice with concrete examples: lottery tickets, committee selections, card hands. These tangible scenarios make the abstract concept click. Once you internalize the "choose k from n" pattern, you'll spot it everywhere in math and real life.