Combination Calculator

What is a Combination Calculator?

A combination calculator computes the number of ways to select r objects from n total objects when the order of selection doesn't matter. Denoted as C(n,r), nCr, or (n choose r), combinations are calculated using the formula C(n,r) = n!/(r!(n-r)!), representing the binomial coefficient. For example, C(5,3) = 5!/(3!×2!) = 120/(6×2) = 10, meaning there are 10 ways to select 3 objects from 5 when order doesn't matter. Combinations are fundamental to probability theory, statistics, binomial distributions, and countless scenarios where selection order is irrelevant. The critical distinction between combinations and permutations: combinations count unordered selections (selecting ABC, BAC, and CAB are all the same), while permutations count ordered arrangements (ABC, BAC, and CAB are different). This calculator handles standard combinations, provides detailed step-by-step solutions, and explains applications across lottery analysis, committee formation, hand dealt in card games, quality control sampling, and statistical hypothesis testing. Understanding combinations is essential for calculating probabilities, analyzing sampling methods, working with binomial and hypergeometric distributions, and solving real-world selection problems where order is irrelevant. The combination formula appears throughout mathematics: Pascal's triangle consists entirely of binomial coefficients, the binomial theorem uses combinations to expand (a+b)^n, and many probability distributions involve combinatorial terms in their formulas.

Key Features

Standard Combination Calculation

Calculate C(n,r) = n!/(r!(n-r)!) for any valid n and r instantly

Step-by-Step Solutions

See complete calculation process with factorial expansions and cancellations

Pascal's Triangle Display

Visualize combinations in Pascal's triangle to understand patterns and relationships

Binomial Coefficient Properties

Explore symmetry property C(n,r) = C(n,n-r) and other relationships

Large Number Support

Handle calculations with large n and r using optimized algorithms

Related Calculations

Compare with permutations and understand P(n,r) = C(n,r) × r! relationship

Probability Applications

Apply combinations to lottery odds, card games, and probability problems

Free & Instant

No registration required with immediate calculation results

How to Use the Combination Calculator

Enter Total Objects (n)

Input the total number of objects available. For example, for a lottery with 49 numbers, enter n = 49.

Enter Objects to Select (r)

Input how many objects you're selecting. For choosing 6 lottery numbers from 49, enter r = 6.

View Combination Result

See C(n,r) calculated instantly. For C(49,6) = 49!/(6!×43!) = 13,983,816 possible selections.

Review Calculation Steps

Examine the detailed solution showing factorial calculations, cancellations, and simplifications used.

Check Symmetry Property

Verify that C(49,6) = C(49,43), demonstrating the symmetry of binomial coefficients.

Apply to Your Problem

Use the result in your lottery odds analysis, sampling calculation, or probability problem.

Combination Tips

Order Doesn't Matter in Combinations: Use combinations when selection order is irrelevant (ABC = BAC = CAB). Use permutations when order matters.
Use Symmetry for Efficiency: If r > n/2, calculate C(n,n-r) instead - it's computationally easier. C(100,97) = C(100,3) is much simpler.
Cancel Factorials Early: For C(n,r) = n!/(r!(n-r)!), cancel (n-r)! from numerator and denominator before multiplying to avoid huge numbers.
Remember Special Cases: C(n,0) = 1, C(n,1) = n, C(n,n) = 1, and C(n,r) = 0 if r > n.
Check Your Constraint: Verify 0 ≤ r ≤ n (can't select more objects than available). If r > n, the combination is zero or undefined.
Pascal's Triangle Pattern: Each entry equals the sum of the two entries above it: C(n,r) = C(n-1,r-1) + C(n-1,r).

Frequently Asked Questions

What is a combination and how does it differ from a permutation?

A combination is an unordered selection of objects, while a permutation is an ordered arrangement. This fundamental difference determines which formula applies to your problem. In combinations, order doesn't matter: selecting persons A, B, C for a committee is the same regardless of selection order (ABC = BAC = CAB - all represent the same committee). The combination formula C(n,r) = n!/(r!(n-r)!) counts these unordered selections. In permutations, order matters: arranging A, B, C for president, VP, secretary creates different outcomes (ABC ≠ BAC ≠ CAB). The permutation formula P(n,r) = n!/(n-r)! counts ordered arrangements. The mathematical relationship: P(n,r) = C(n,r) × r!, because each combination can be arranged in r! different orders. Example: From 10 people, selecting any 3 for a committee is C(10,3) = 120 combinations. But selecting president, VP, and treasurer (order matters) is P(10,3) = 720 permutations. Note: 720 = 120 × 6, where 6 = 3! is the number of ways to order 3 positions. Use combinations for: lottery numbers (order irrelevant), committee selection, card hands (poker hand doesn't depend on dealing order), survey sampling. Use permutations for: race podium positions, password characters, seating arrangements, tournament brackets. The key question: does rearranging the selected objects create a different outcome? If yes, use permutations. If no, use combinations.

How do you calculate combinations using the formula C(n,r) = n!/(r!(n-r)!)?

The combination formula C(n,r) = n!/(r!(n-r)!) calculates unordered selections by dividing n! by both r! and (n-r)!. Let's calculate C(6,2) step by step: C(6,2) = 6!/(2!×(6-2)!) = 6!/(2!×4!). Expand factorials: 6! = 720, 2! = 2, 4! = 24. Calculate: 720/(2×24) = 720/48 = 15. So there are 15 ways to select 2 objects from 6. For efficiency, use cancellation: C(6,2) = (6×5×4!)/(2!×4!) = (6×5)/2 = 30/2 = 15. The 4! cancels immediately. General approach: write out enough terms of n! to cancel (n-r)!, leaving r consecutive terms from n: C(n,r) = (n×(n-1)×...×(n-r+1))/(r×(r-1)×...×1). For C(10,3): numerator has 3 terms from 10: (10×9×8), denominator is 3! = 6, giving (10×9×8)/6 = 720/6 = 120. Why this formula? Start with P(n,r) = n!/(n-r)! permutations (ordered arrangements), then divide by r! because the r selected objects can be arranged in r! ways, but these arrangements represent the same combination. So C(n,r) = P(n,r)/r! = [n!/(n-r)!]/r! = n!/(r!(n-r)!). Special cases: C(n,0) = 1 (one way to select nothing), C(n,1) = n (n ways to select one object), C(n,n) = 1 (one way to select all objects), C(n,r) = C(n,n-r) (symmetry property - selecting r objects is equivalent to leaving (n-r) objects). This formula is the foundation of binomial coefficients, appearing throughout probability, statistics, and algebra.

What is Pascal's Triangle and how does it relate to combinations?

Pascal's Triangle is a triangular array of numbers where each number is the sum of the two numbers directly above it, and remarkably, the nth row contains the binomial coefficients C(n,r) for r = 0 to n. The triangle begins: 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; Row 5: 1 5 10 10 5 1. Each entry in row n, position r (counting from 0) equals C(n,r). For example, row 5 contains: C(5,0)=1, C(5,1)=5, C(5,2)=10, C(5,3)=10, C(5,4)=5, C(5,5)=1. The triangle's construction rule reflects the combination property: C(n,r) = C(n-1,r-1) + C(n-1,r), which has a beautiful combinatorial interpretation - to select r objects from n, either include a specific object (then choose r-1 from remaining n-1) or exclude it (choose r from remaining n-1). Pascal's Triangle reveals many patterns: rows sum to 2^n (total subsets of n objects), diagonals contain natural numbers, triangular numbers, tetrahedral numbers, and Fibonacci numbers appear in diagonal sums. The symmetry C(n,r) = C(n,n-r) is visually obvious (triangle is symmetric). Historical note: though named for Blaise Pascal (1653), the triangle was known centuries earlier in China, India, and Persia. Applications extend beyond combinations: binomial expansion coefficients (a+b)^n, probability distributions, cellular automaton rules (Sierpinski triangle emerges from Pascal's Triangle mod 2), and polynomial interpolation. Understanding Pascal's Triangle provides intuition for combinatorial identities and connects discrete mathematics to number theory, algebra, and probability.

How are combinations used in lottery odds and gambling?

Combinations calculate lottery odds because lottery numbers are unordered selections - drawing numbers 4, 15, 23, 42, 8, 16 is the same as drawing 8, 15, 16, 23, 4, 42. For a lottery choosing r numbers from n possible numbers, there are C(n,r) possible combinations, each equally likely. Popular example: 6/49 lottery selects 6 numbers from 49, giving C(49,6) = 49!/(6!×43!) = 13,983,816 possible combinations. Your ticket represents 1 combination, so the probability of winning the jackpot is 1/13,983,816 ≈ 0.0000000715 or about 1 in 14 million. For Powerball (select 5 from 69 white balls and 1 from 26 red balls): C(69,5) × C(26,1) = 11,238,513 × 26 = 292,201,338 possible combinations, giving jackpot odds of about 1 in 292 million. In poker, calculating hand probabilities uses combinations: a 5-card hand from 52 cards has C(52,5) = 2,598,960 possible hands. Probability of a specific four-of-a-kind (say, four aces): C(4,4) ways to get four aces × C(48,1) ways to choose the fifth card = 1 × 48 = 48 hands, giving probability 48/2,598,960 ≈ 0.0000185. For a flush (5 cards same suit): C(4,1) ways to choose suit × C(13,5) ways to choose 5 cards from that suit = 4 × 1,287 = 5,148 hands, probability ≈ 0.00198. Understanding combinations enables calculating exact probabilities in games of chance, explaining why casinos have house advantage, and revealing why lottery is a poor investment (negative expected value despite occasional large payouts).

What is the symmetry property of combinations and why is it useful?

The symmetry property states that C(n,r) = C(n,n-r), meaning selecting r objects from n is equivalent to leaving (not selecting) n-r objects. Mathematically, this follows from the formula: C(n,r) = n!/(r!(n-r)!) and C(n,n-r) = n!/((n-r)!r!), which are identical (multiplication is commutative). For example, C(10,3) = C(10,7) = 120 - choosing 3 people from 10 is equivalent to choosing 7 to not be on the committee. This symmetry is visually evident in Pascal's Triangle (each row is symmetric). The property is computationally useful: when calculating C(n,r) with r > n/2, compute C(n,n-r) instead for efficiency. For C(49,45), compute C(49,4) = (49×48×47×46)/(4×3×2×1) = 211,876, much simpler than working with 45!. This also explains why C(n,0) = C(n,n) = 1 (selecting all objects is the same as selecting none to exclude). The symmetry has deep combinatorial meaning: every selection corresponds to a complementary exclusion. In probability, this creates useful relationships: P(at least r successes) = P(at most n-r failures). In binomial distribution, probabilities are symmetric around the mean for p=0.5. In graph theory, the number of r-element subsets equals the number of (n-r)-element subsets. Understanding this symmetry simplifies calculations, reveals mathematical elegance, and provides alternative problem perspectives. It's a beautiful example of how algebraic formulas reflect combinatorial reality.

How do you use combinations to calculate probabilities in sampling?

Combinations are essential for calculating probabilities in sampling without replacement, where you select items without repeating selections. The basic probability formula: P(event) = (favorable outcomes)/(total outcomes), where both terms typically involve combinations. Example: A box contains 30 parts (25 good, 5 defective). If you randomly select 4 parts, what's the probability of getting exactly 2 defective? Total ways to select 4 from 30: C(30,4) = 27,405. Favorable outcomes: select 2 defective from 5 AND 2 good from 25: C(5,2) × C(25,2) = 10 × 300 = 3,000. Probability: 3,000/27,405 ≈ 0.1095 or about 11%. This follows the hypergeometric distribution for sampling without replacement. For sampling with replacement (items returned after selection), use different formulas. In quality control, acceptance sampling uses combinations: accept a lot if at most 1 defect in a sample of 10. If the lot has 5% defects (50 defects in 1000 items), probability calculations determine acceptance rates. In genetics, combinations calculate genotype probabilities: selecting 2 alleles from available variants. In survey sampling, combinations determine the number of possible samples: selecting 100 people from 10,000 population gives C(10,000,100) possible samples (astronomically large, justifying sampling theory). The key insight: when sampling without replacement from a finite population with k successes and n-k failures, the probability of exactly r successes in a sample of size m is: P(X=r) = [C(k,r) × C(n-k,m-r)]/C(n,m). Mastering combination-based probability enables analyzing surveys, quality control, card games, and any scenario involving selection from finite populations.

What are some real-world applications of combinations?

Combinations have extensive real-world applications across diverse fields where selection order doesn't matter. In lottery and gambling, combinations calculate odds: lottery tickets, poker hands, raffles all use C(n,r) to determine probabilities and payouts. In committee formation and team selection, choosing r people from n available uses combinations: selecting a 5-person committee from 20 employees gives C(20,5) = 15,504 possible committees. In quality control and sampling, selecting items to inspect from a batch uses combinations: testing 10 units from 100 manufactured items gives C(100,10) possible samples, crucial for statistical quality control. In clinical trials, selecting patients for treatment vs control groups uses combinations to determine possible allocations. In portfolio management, selecting r stocks from n available stocks to form a portfolio uses C(n,r): choosing 5 stocks from 50 gives C(50,5) = 2,118,760 possible portfolios. In menu planning, selecting r dishes from n options creates combination-based meal plans. In tournament scheduling, selecting pairs of teams for matches uses C(n,2) = n(n-1)/2: 8 teams playing all pairs requires C(8,2) = 28 matches. In genetics, combinations of alleles determine genotypes: selecting 2 alleles from variants. In network theory, selecting r nodes from n nodes to analyze connectivity or form subgraphs uses combinations. In survey design, selecting questions from question banks uses C(n,r). In music, selecting r notes from n available notes (ignoring order) creates unique chord structures. Understanding combinations enables counting selections, calculating probabilities, analyzing sampling methods, and solving optimization problems across business, science, engineering, and everyday decision-making.

What is the relationship between combinations and the binomial theorem?

Combinations form the coefficients in the binomial theorem, which states: (a+b)^n = Σ C(n,k) × a^(n-k) × b^k for k from 0 to n. The binomial coefficients C(n,k) determine how many terms contribute to each power combination in the expansion. For example, (a+b)³ = C(3,0)a³b⁰ + C(3,1)a²b¹ + C(3,2)a¹b² + C(3,3)a⁰b³ = 1a³ + 3a²b + 3ab² + 1b³. The coefficients 1,3,3,1 come from row 3 of Pascal's Triangle, and equal C(3,0), C(3,1), C(3,2), C(3,3). Why combinations appear: when expanding (a+b)^n = (a+b)(a+b)...(a+b) [n times], each term in the result comes from choosing either 'a' or 'b' from each factor. A term with k b's (and n-k a's) arises from choosing 'b' from k factors and 'a' from the remaining n-k factors. The number of ways to choose which k factors contribute 'b' is C(n,k), explaining the coefficient. Setting a=b=1 gives: 2^n = Σ C(n,k), showing that the sum of all binomial coefficients in row n equals 2^n (total subsets of n elements). Setting a=1, b=-1 gives: 0 = Σ (-1)^k C(n,k), showing alternating sum equals zero (for n≥1). The binomial theorem has extensive applications: calculating powers efficiently, approximating (1+x)^n for small x (binomial approximation), generating function theory, probability distributions (binomial distribution uses these coefficients), and combinatorial identities. Understanding the combination-binomial theorem connection reveals why Pascal's Triangle appears in algebra, probability, and number theory, demonstrating deep unity in mathematics.

Why Use Our Combination Calculator?

Calculating combinations should be simple and insightful. Our combination calculator computes C(n,r) instantly while providing step-by-step solutions, factorial simplifications, and clear explanations. Whether you're calculating lottery odds, determining sample sizes, solving probability problems, or working with binomial coefficients, our tool delivers accurate results with educational content. With support for large numbers and visualization of Pascal's Triangle patterns, you get both computational power and mathematical understanding.

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