Combinations & Permutations Calculator
Calculate combinations (nCr) and permutations (nPr), with and without repetition. Explore how counts grow as n and r change.
Calculate combinations (nCr) and permutations (nPr), with and without repetition. Explore how counts grow as n and r change.
Combinations and permutations are fundamental counting methods in combinatorics, the branch of mathematics that studies counting, arrangement, and selection problems. These concepts are essential for solving probability problems, calculating lottery odds, analyzing password security, and understanding many real-world scenarios involving selection and arrangement. This tool helps you calculate combinations (nCr) and permutations (nPr) with or without repetition, providing formulas, results, and visualizations for counting problems. Whether you're a student learning combinatorics, a researcher analyzing probability distributions, a data analyst calculating odds, or a business professional evaluating security configurations, understanding combinations and permutations enables you to solve counting problems, calculate probabilities, and make informed decisions based on combinatorial analysis.
For students and researchers, this tool demonstrates practical applications of combinatorics, factorial calculations, and counting principles. The combinations and permutations calculations show how factorials, multiplicative principles, and counting formulas combine to produce the number of ways to select or arrange items. Students can use this tool to verify homework calculations, understand how different counting modes (with/without repetition, order matters/doesn't matter) address different problems, and explore concepts like factorial growth, binomial coefficients, and the relationship between combinations and permutations. Researchers can apply combinations and permutations to analyze probability distributions, calculate binomial coefficients, evaluate password spaces, and understand counting principles in various fields.
For business professionals and practitioners, combinations and permutations provide essential tools for decision-making and risk analysis. Security professionals use permutations to calculate password space sizes, evaluate encryption key strengths, and assess security configurations. Lottery and gaming professionals use combinations to calculate odds, evaluate game fairness, and design lottery systems. Data analysts use combinations and permutations to analyze sampling methods, calculate probability distributions, and evaluate experimental designs. Operations managers use permutations to analyze scheduling possibilities, evaluate arrangement options, and assess operational configurations. Quality control engineers use combinations to analyze sampling plans, calculate defect probabilities, and evaluate quality inspection methods.
For the common person, this tool answers practical counting questions: How many ways can I choose 5 cards from a deck? How many different PIN codes are possible? What are my lottery odds? The tool calculates combinations and permutations for any counting scenario, providing formulas, results, and interpretations. Taxpayers and budget-conscious individuals can use combinations and permutations to understand lottery odds, evaluate password security, calculate probability scenarios, and make informed decisions based on combinatorial analysis. These concepts help you understand the mathematics behind everyday counting problems and probability scenarios.
The fundamental distinction between combinations and permutations is whether order matters. In combinations, selecting items A, B, C is the same as selecting C, B, A—only the group matters, not the arrangement. Use combinations when selecting a committee, choosing lottery numbers, or picking a poker hand. In permutations, ABC and CBA are different because the sequence matters. Use permutations when arranging books on a shelf, assigning race places, or creating ordered sequences. The relationship is: P(n, r) = C(n, r) × r!, showing that permutations count all possible orderings of combinations. For example, C(5, 3) = 10 (groups of 3 from 5), while P(5, 3) = 60 (ordered arrangements of 3 from 5). The ratio is 3! = 6, representing the number of ways to arrange 3 items.
Combinations without repetition count the number of ways to choose r items from n distinct items where order doesn't matter and each item can only be selected once. The formula is C(n, r) = n! / (r!(n-r)!) = n! / (r! × (n-r)!). This is also called "n choose r" or the binomial coefficient. Examples include: choosing 5 cards from a 52-card deck (C(52, 5) = 2,598,960), selecting 3 students from 10 for a committee (C(10, 3) = 120), or picking 6 lottery numbers from 49 (C(49, 6) = 13,983,816). The calculation uses multiplicative simplification: C(n, r) = (n × (n-1) × ... × (n-r+1)) / (r × (r-1) × ... × 1), which is more efficient than computing factorials directly. Special cases: C(n, 0) = 1 (one way to choose nothing), C(n, n) = 1 (one way to choose everything), C(n, r) = C(n, n-r) (symmetry property).
Permutations without repetition count the number of ways to arrange r items from n distinct items in order, where each item can only be used once. The formula is P(n, r) = n! / (n-r)! = n × (n-1) × (n-2) × ... × (n-r+1). This represents the number of ordered arrangements. Examples include: arranging 3 books on a shelf from 10 books (P(10, 3) = 720), assigning 1st, 2nd, 3rd place from 10 contestants (P(10, 3) = 720), or creating a 4-digit code from digits 0-9 without repetition (P(10, 4) = 5,040). Special cases: P(n, 0) = 1 (one way to arrange nothing), P(n, n) = n! (full arrangement of all n items), P(n, 1) = n (n ways to choose one item). The relationship to combinations is: P(n, r) = C(n, r) × r!, showing that permutations equal combinations multiplied by the number of ways to arrange r items.
Combinations with repetition count the number of ways to choose r items from n types when repetition is allowed and order doesn't matter. The formula is C(n+r-1, r) = (n+r-1)! / (r!(n-1)!). This uses the "stars and bars" method, which counts ways to distribute r identical items into n distinct bins. Examples include: choosing 5 donuts from 4 flavors (C(4+5-1, 5) = C(8, 5) = 56), selecting 10 items from 3 categories (C(3+10-1, 10) = C(12, 10) = 66), or distributing 20 identical objects into 5 boxes (C(5+20-1, 20) = C(24, 20) = 10,626). The stars and bars method visualizes the problem as arranging r stars (items) and n-1 bars (dividers between categories) in a line. This formula is useful when items can be reused and only the count in each category matters, not the order.
Permutations with repetition count the number of ways to arrange r items where each position can be any of n distinct items and repetition is allowed. The formula is n^r (n raised to the power of r). This represents the number of ordered arrangements when items can be reused. Examples include: creating a 4-digit PIN using digits 0-9 (10^4 = 10,000), forming a 3-letter code from 26 letters (26^3 = 17,576), or generating a 6-character password from 62 characters (62^6 = 56,800,235,584). This is the simplest counting formula and grows exponentially with r. Special cases: n^0 = 1 (one way to arrange nothing), n^1 = n (n ways to choose one item). Permutations with repetition are used extensively in password security analysis, encryption key space calculations, and scenarios where each position can independently be any of n options.
The factorial of n (denoted n!) is the product of all positive integers from 1 to n: n! = n × (n-1) × (n-2) × ... × 2 × 1. Factorials grow extremely fast: 5! = 120, 10! = 3,628,800, 15! = 1,307,674,368,000, 20! ≈ 2.43 × 10^18. By convention, 0! = 1, which makes formulas consistent and ensures C(n, 0) = 1 (one way to choose nothing). Factorials are used in combinations and permutations formulas, but the calculations use multiplicative simplification to avoid computing large factorials directly. Properties: n! = n × (n-1)!, (n+1)! = (n+1) × n!, and n! / (n-r)! = n × (n-1) × ... × (n-r+1) for permutations. The rapid growth of factorials is why combinatorial problems quickly produce astronomically large numbers, requiring careful calculation methods and scientific notation for display.
In combinations and permutations formulas, n represents the total number of distinct items available to choose from, and r represents how many items you're selecting or arranging. For example, choosing 5 cards from a 52-card deck means n = 52 and r = 5. The constraint r ≤ n applies to combinations and permutations without repetition (you can't select more items than available), but r can exceed n when repetition is allowed (you can select more items than types available). Both n and r must be non-negative integers. Special cases: when r = 0, C(n, 0) = 1 and P(n, 0) = 1 (one way to choose/arrange nothing); when r = n (for no-repetition), C(n, n) = 1 and P(n, n) = n! (one way to choose everything, n! ways to arrange everything).
To choose the correct counting method, ask two questions: (1) Does order matter? If yes → Permutation; if no → Combination. (2) Can items be repeated? If yes → With Repetition; if no → Without Repetition. This gives four modes: (a) Combinations Without Repetition—order doesn't matter, no repetition (e.g., choosing lottery numbers), (b) Permutations Without Repetition—order matters, no repetition (e.g., arranging books on a shelf), (c) Combinations With Repetition—order doesn't matter, repetition allowed (e.g., choosing donuts from flavors), (d) Permutations With Repetition—order matters, repetition allowed (e.g., creating PIN codes). Always identify whether order matters and whether repetition is allowed before selecting the counting method.
First, determine whether the order of selection or arrangement matters. If order matters (e.g., arranging books, assigning places, creating sequences), you need permutations. If order doesn't matter (e.g., choosing a committee, selecting lottery numbers, picking a hand), you need combinations. This is the most important decision, as it determines whether you use C(n, r) or P(n, r) formulas. Think about whether ABC is different from CBA in your scenario—if yes, use permutations; if no, use combinations.
Next, determine whether items can be repeated. If each item can only be selected once (e.g., dealing cards, choosing team members), use "Without Repetition". If items can be reused (e.g., digits in a PIN, flavors in ice cream scoops), use "With Repetition". This determines whether you use the standard formulas (C(n, r) or P(n, r)) or the repetition formulas (C(n+r-1, r) or n^r). Think about whether the same item can appear multiple times in your selection.
Enter the total number of distinct items (n) and the number of items to select or arrange (r). For example, if choosing 5 cards from a 52-card deck, enter n = 52 and r = 5. If creating a 4-digit PIN from digits 0-9, enter n = 10 and r = 4. Make sure n and r are non-negative integers. For combinations and permutations without repetition, r cannot exceed n. For combinations and permutations with repetition, r can exceed n. The tool validates inputs and shows errors if values are invalid.
Select the appropriate counting mode based on your answers to steps 1 and 2: "Combinations Without Repetition" (order doesn't matter, no repetition), "Permutations Without Repetition" (order matters, no repetition), "Combinations With Repetition" (order doesn't matter, repetition allowed), or "Permutations With Repetition" (order matters, repetition allowed). The tool will use the appropriate formula for your selected mode. Make sure your selection matches your problem scenario.
Click "Calculate" or submit the form to compute the combinations or permutations. The tool displays the calculated value, the formula used, and an interpretation summary explaining what the result means. For large values, the tool displays results in scientific notation or with appropriate formatting. Review the interpretation summary to understand what the number represents in your specific scenario. The tool also shows the formula breakdown, helping you understand how the calculation was performed.
Optionally, generate a table showing values for different r values (or different n values) to see how the count changes. This helps you understand the relationship between parameters and results. For example, you can see how C(52, r) changes as r increases from 1 to 10, or how P(10, r) changes for different r values. The table and chart visualization help you understand the combinatorial function's behavior and identify patterns in the counting results.
Combinations without repetition use the binomial coefficient formula:
Formula: C(n, r) = n! / (r!(n-r)!)
Multiplicative Method (more efficient):
C(n, r) = (n × (n-1) × ... × (n-r+1)) / (r × (r-1) × ... × 1)
Optimization: Use k = min(r, n-r) for efficiency
Special Cases: C(n, 0) = 1, C(n, n) = 1, C(n, r) = C(n, n-r)
The multiplicative method avoids computing large factorials directly by canceling common factors. For example, C(52, 5) = (52 × 51 × 50 × 49 × 48) / (5 × 4 × 3 × 2 × 1) = 2,598,960. The optimization uses C(n, r) = C(n, n-r) to minimize the number of multiplications (use the smaller of r and n-r). The calculation checks for overflow and rounds the result to handle floating-point precision issues. This method is more efficient and numerically stable than computing factorials separately.
Permutations without repetition use the falling factorial formula:
Formula: P(n, r) = n! / (n-r)!
Multiplicative Method:
P(n, r) = n × (n-1) × (n-2) × ... × (n-r+1)
Special Cases: P(n, 0) = 1, P(n, n) = n!, P(n, 1) = n
The multiplicative method directly computes the product of r consecutive integers starting from n, avoiding factorial calculations. For example, P(10, 3) = 10 × 9 × 8 = 720. This is more efficient than computing 10! / 7! = 3,628,800 / 5,040 = 720. The calculation checks for overflow and handles large values appropriately. The relationship to combinations is: P(n, r) = C(n, r) × r!, showing that permutations equal combinations multiplied by the number of ways to arrange r items.
Combinations with repetition use the stars and bars method:
Formula: C(n+r-1, r) = (n+r-1)! / (r!(n-1)!)
Stars and Bars Method:
Count ways to arrange r stars (items) and n-1 bars (dividers)
Example: C(4+5-1, 5) = C(8, 5) = 56 ways to choose 5 donuts from 4 flavors
The stars and bars method visualizes the problem as arranging r identical stars (representing items) and n-1 bars (representing dividers between categories) in a line. The number of arrangements equals C(n+r-1, r) = C(n+r-1, n-1). For example, choosing 5 donuts from 4 flavors: arrange 5 stars and 3 bars (4-1=3) in 8 positions, giving C(8, 5) = 56 ways. This formula is calculated using the standard combination function with adjusted parameters: combination(n + r - 1, r).
Permutations with repetition use the simplest counting formula:
Formula: n^r (n raised to the power of r)
Calculation: Multiply n by itself r times
Special Cases: n^0 = 1, n^1 = n
Example: 10^4 = 10,000 ways to create a 4-digit PIN
Permutations with repetition are calculated using exponentiation: n^r. This represents the number of ordered arrangements when each of r positions can independently be any of n options. For example, a 4-digit PIN using digits 0-9: each of 4 positions can be any of 10 digits, giving 10^4 = 10,000 possible PINs. The calculation uses Math.pow(n, r) and checks for overflow. This formula grows exponentially with r, making it important for password security analysis and encryption key space calculations.
Let's calculate the number of ways to choose 6 numbers from 49 for a lottery:
Given: n = 49, r = 6, mode = Combinations Without Repetition
Step 1: Identify the Formula
C(49, 6) = 49! / (6!(49-6)!) = 49! / (6! × 43!)
Step 2: Use Multiplicative Method
C(49, 6) = (49 × 48 × 47 × 46 × 45 × 44) / (6 × 5 × 4 × 3 × 2 × 1)
= 10,068,347,520 / 720 = 13,983,816
Step 3: Interpret the Result
There are 13,983,816 different ways to choose 6 numbers from 49.
Step 4: Calculate Odds
If you buy one ticket, your odds of winning the jackpot are 1 in 13,983,816.
Interpretation:
This demonstrates why lottery jackpots are so rare—there are nearly 14 million different number combinations, and you only have one ticket. The probability of winning is 1/13,983,816 ≈ 0.00000715% (about 7 in 100 million).
This example demonstrates how combinations without repetition calculate lottery odds. The multiplicative method avoids computing 49! directly, which would be astronomically large. Instead, we compute the product of 6 consecutive integers starting from 49 and divide by 6!. The result shows there are nearly 14 million different ways to choose 6 numbers, explaining why lottery jackpots are so rare. This same principle applies to any scenario where you're choosing a subset from a larger set without regard to order.
Let's calculate the number of possible 4-digit PIN codes using digits 0-9:
Given: n = 10 (digits 0-9), r = 4, mode = Permutations With Repetition
Step 1: Identify the Formula
P(10, 4) with repetition = 10^4
Step 2: Calculate
10^4 = 10 × 10 × 10 × 10 = 10,000
Step 3: Compare to Without Repetition
P(10, 4) without repetition = 10 × 9 × 8 × 7 = 5,040
Step 4: Interpret Security Implications
With repetition: 10,000 possible PINs (allows 1111, 2222, etc.)
Without repetition: 5,040 possible PINs (no repeated digits)
Interpretation:
A 4-digit PIN with repetition has 10,000 possible combinations, meaning an attacker would need to try up to 10,000 combinations to guess it (on average, 5,000 attempts). Without repetition, there are only 5,040 possibilities, making it slightly less secure. This demonstrates why longer PINs or passwords are more secure—they exponentially increase the number of possibilities.
This example demonstrates how permutations with repetition calculate password/PIN space sizes. The formula n^r shows exponential growth—each additional position multiplies the possibilities by n. A 4-digit PIN has 10^4 = 10,000 possibilities, while a 6-digit PIN has 10^6 = 1,000,000 possibilities. This exponential growth is why longer passwords are exponentially more secure. The comparison with without-repetition shows how restrictions reduce the space, though the difference is less dramatic for small r values.
A student needs to choose 3 students from 10 for a committee. Order doesn't matter, and each student can only be chosen once. Using the tool with n=10, r=3, mode="Combinations Without Repetition", the tool calculates C(10, 3) = 120. The student learns that there are 120 different ways to choose 3 students from 10. The formula C(10, 3) = 10! / (3! × 7!) = (10 × 9 × 8) / (3 × 2 × 1) = 720 / 6 = 120 shows how the calculation works. This helps them understand that the order of selection doesn't matter in committee formation.
A person wants to calculate their odds of winning a 6/49 lottery. Order doesn't matter, and each number can only appear once. Using the tool with n=49, r=6, mode="Combinations Without Repetition", the tool calculates C(49, 6) = 13,983,816. The person learns that there are 13,983,816 different ways to choose 6 numbers, meaning their odds of winning with one ticket are 1 in 13,983,816 (approximately 0.00000715%). This helps them understand why lottery jackpots are so rare and why buying multiple tickets only slightly improves odds.
A security analyst evaluates a 6-character password using 62 characters (26 lowercase + 26 uppercase + 10 digits). Each position can be any character, and repetition is allowed. Using the tool with n=62, r=6, mode="Permutations With Repetition", the tool calculates 62^6 = 56,800,235,584. The analyst learns that there are over 56 billion possible passwords, meaning an attacker would need to try up to 56 billion combinations (on average, 28 billion attempts) to guess it. This demonstrates strong password security for a 6-character password with a large character set.
A person wants to choose 5 donuts from 4 flavors, where they can pick multiple donuts of the same flavor. Order doesn't matter (only the count of each flavor matters), and repetition is allowed. Using the tool with n=4, r=5, mode="Combinations With Repetition", the tool calculates C(4+5-1, 5) = C(8, 5) = 56. The person learns that there are 56 different ways to choose 5 donuts from 4 flavors when repetition is allowed. This uses the stars and bars method, which counts ways to distribute 5 identical items (donuts) into 4 distinct categories (flavors).
A bookstore manager wants to arrange 3 books on a display shelf from 10 available books. Order matters (different arrangements create different displays), and each book can only appear once. Using the tool with n=10, r=3, mode="Permutations Without Repetition", the tool calculates P(10, 3) = 10 × 9 × 8 = 720. The manager learns that there are 720 different ways to arrange 3 books on the shelf. This helps them understand the number of display options available and plan shelf arrangements effectively.
A researcher analyzes poker hand probabilities. A 5-card hand from a 52-card deck: order doesn't matter (a hand is a hand regardless of card order), and each card can only appear once. Using the tool with n=52, r=5, mode="Combinations Without Repetition", the tool calculates C(52, 5) = 2,598,960. The researcher learns that there are 2,598,960 different possible 5-card hands. This is the denominator for calculating specific poker hand probabilities (e.g., probability of a royal flush = 4 / 2,598,960 ≈ 0.000154%).
A user compares C(10, 3) = 120 and P(10, 3) = 720. They learn that P(10, 3) = C(10, 3) × 3! = 120 × 6 = 720. This demonstrates that permutations count all possible orderings of combinations. For every combination of 3 items, there are 3! = 6 ways to arrange them, so permutations = combinations × r!. This relationship helps users understand why permutations are always greater than or equal to combinations (when r > 1), and why the difference grows with r. The formula P(n, r) = C(n, r) × r! is fundamental to understanding the relationship between these counting methods.
Don't confuse combinations and permutations—the key difference is whether order matters. Use combinations when order doesn't matter (e.g., choosing a committee, selecting lottery numbers, picking a hand). Use permutations when order matters (e.g., arranging books, assigning places, creating sequences). If you're unsure, ask: "Is ABC different from CBA?" If yes, use permutations; if no, use combinations. This is the most common mistake in combinatorics problems, and choosing the wrong method will give incorrect results.
Don't use the wrong repetition mode—this will give incorrect results. Use "Without Repetition" when each item can only be selected once (e.g., dealing cards, choosing team members, selecting lottery numbers). Use "With Repetition" when items can be reused (e.g., digits in a PIN, flavors in ice cream scoops, characters in a password). If you're unsure, ask: "Can the same item appear multiple times?" If yes, use with repetition; if no, use without repetition. Using the wrong mode can dramatically change the result.
For combinations and permutations without repetition, r cannot exceed n (you can't select more items than available). However, for combinations and permutations with repetition, r can exceed n (you can select more items than types available). Don't assume r must always be ≤ n—this only applies to no-repetition modes. For example, choosing 10 donuts from 4 flavors (with repetition) is valid: C(4+10-1, 10) = C(13, 10) = 286. But choosing 10 cards from a 4-card deck (without repetition) is invalid: C(4, 10) is undefined.
Don't compute factorials directly for large numbers—this causes overflow and numerical errors. Use multiplicative simplification instead. For combinations: C(n, r) = (n × (n-1) × ... × (n-r+1)) / (r × (r-1) × ... × 1). For permutations: P(n, r) = n × (n-1) × ... × (n-r+1). These methods avoid computing large factorials and are more efficient and numerically stable. The tool uses these multiplicative methods automatically, but understanding them helps you verify calculations and avoid errors in manual work.
Don't forget special cases: C(n, 0) = 1 (one way to choose nothing), C(n, n) = 1 (one way to choose everything), P(n, 0) = 1 (one way to arrange nothing), P(n, n) = n! (n! ways to arrange everything), n^0 = 1 (one way to arrange nothing with repetition), and 0! = 1 (by convention). These special cases are important for edge cases and make formulas consistent. Always verify that your calculation handles these cases correctly, especially when r = 0 or r = n.
Don't misapply the stars and bars formula for combinations with repetition. The formula C(n+r-1, r) applies when: (1) order doesn't matter, (2) repetition is allowed, and (3) you're choosing r items from n types. This is different from standard combinations C(n, r), which requires r ≤ n. The stars and bars method counts ways to distribute r identical items into n distinct bins, which is equivalent to choosing r items from n types with repetition. Make sure you understand when to use this formula versus standard combinations.
Permutations with repetition (n^r) grow exponentially with r, which is much faster than combinations or permutations without repetition. Don't underestimate how quickly n^r grows—each additional position multiplies possibilities by n. For example, 10^4 = 10,000, but 10^6 = 1,000,000 (100 times larger). This exponential growth is why longer passwords are exponentially more secure. Always consider the exponential nature when evaluating security or probability scenarios involving permutations with repetition.
Always use the decision tree to choose the correct counting method: (1) Does order matter? If yes → Permutation; if no → Combination. (2) Can items be repeated? If yes → With Repetition; if no → Without Repetition. This gives you the correct mode: Combinations Without Repetition, Permutations Without Repetition, Combinations With Repetition, or Permutations With Repetition. Following this decision tree ensures you use the correct formula and get accurate results. Don't guess—systematically work through the decision tree for each problem.
Remember that P(n, r) = C(n, r) × r!, showing that permutations count all possible orderings of combinations. For every combination of r items, there are r! ways to arrange them, so permutations = combinations × r!. This relationship helps you verify calculations and understand why permutations are always greater than or equal to combinations (when r > 1). For example, C(10, 3) = 120 and P(10, 3) = 720, and indeed 120 × 6 = 720. Use this relationship to check your work and understand the connection between these counting methods.
For large n and r, use multiplicative methods instead of computing factorials directly. For combinations: C(n, r) = (n × (n-1) × ... × (n-r+1)) / (r × (r-1) × ... × 1). For permutations: P(n, r) = n × (n-1) × ... × (n-r+1). These methods avoid computing large factorials and are more efficient and numerically stable. The tool uses these methods automatically, but understanding them helps you verify calculations and avoid errors in manual work. Always use the optimization C(n, r) = C(n, n-r) to minimize multiplications.
For combinations with repetition, understand the stars and bars method: C(n+r-1, r) counts ways to distribute r identical items (stars) into n distinct bins (categories) using n-1 dividers (bars). This visual method helps you understand why the formula is C(n+r-1, r) rather than C(n, r). For example, choosing 5 donuts from 4 flavors: arrange 5 stars and 3 bars in 8 positions, giving C(8, 5) = 56 ways. Understanding this method helps you apply the formula correctly and recognize when to use combinations with repetition.
Remember that permutations with repetition (n^r) grow exponentially with r. Each additional position multiplies possibilities by n, making longer sequences exponentially more numerous. For example, 10^4 = 10,000, 10^5 = 100,000, 10^6 = 1,000,000. This exponential growth is why longer passwords are exponentially more secure. When evaluating security or probability scenarios, always consider the exponential nature of n^r. A small increase in r can dramatically increase the number of possibilities, making systems more secure or probabilities smaller.
Generate tables showing values for different r (or n) to understand how combinations and permutations change with parameters. For example, see how C(52, r) changes as r increases from 1 to 10, or how P(10, r) changes for different r values. The table and chart visualization help you understand the combinatorial function's behavior, identify patterns (like symmetry in combinations: C(n, r) = C(n, n-r)), and see how values grow. This is especially useful for understanding factorial growth and the relationship between parameters and results.
Use combinations and permutations to solve real-world probability problems. For example: (1) Lottery odds = 1 / C(n, r), (2) Poker hand probability = (number of ways to get hand) / C(52, 5), (3) Password security = 1 / n^r (for permutations with repetition), (4) Committee selection probability = 1 / C(n, r). Understanding combinations and permutations enables you to calculate probabilities, evaluate odds, assess security, and solve counting problems in various fields. Always identify the counting method first, then use it to calculate probabilities.
• Distinct Items Assumption: Standard formulas assume all n items are distinguishable. When items are identical or grouped (e.g., arrangements with repeated elements), different formulas apply—multinomial coefficients or adjusted permutation formulas are required for non-distinct items.
• Computational Limits: Factorials grow extremely rapidly (20! ≈ 2.43 × 10¹⁸). For very large n and r values, results may exceed JavaScript's safe integer range (2⁵³) or floating-point precision limits—professional mathematical software provides arbitrary-precision arithmetic for extreme values.
• Model Matching: Choosing the wrong counting mode (combinations vs. permutations, with vs. without repetition) produces incorrect results. Real-world problems often have subtle constraints—carefully analyze whether order matters and whether repetition is allowed in your specific scenario.
• Theoretical vs. Practical: Calculated counts assume idealized conditions. Real-world constraints (physical limitations, dependencies, conditional rules) may reduce the actual number of valid outcomes—always verify that your problem truly matches the mathematical model's assumptions.
Important Note: This calculator is strictly for educational and informational purposes only. It does not provide professional mathematical consulting, gambling advice, security analysis, or probability assessments for real-world applications. Combinatorial calculations are theoretical counts assuming idealized conditions—real problems often have additional constraints not captured by standard formulas. Results should be verified using professional mathematical software (MATLAB, Mathematica, Python SymPy) for any applications involving security analysis, lottery systems, cryptographic key spaces, or professional probability assessments. For critical decisions involving risk, gambling, financial calculations, or security system design, always consult qualified mathematicians or domain experts who can evaluate whether the counting model appropriately matches your specific scenario.
The mathematical formulas and combinatorics concepts used in this calculator are based on established mathematical theory and authoritative academic sources:
Common questions about combinations and permutations, nCr and nPr formulas, factorials, stars and bars method, lottery odds, password security, and how to use this calculator for homework and statistics practice.
The key difference is whether order matters. In combinations, selecting items A, B, C is the same as C, B, A—only the group matters, not the arrangement. In permutations, ABC and CBA are different because the sequence matters. Use combinations for selecting committees, lottery tickets, or poker hands. Use permutations for arranging books on a shelf, assigning race places, or creating ordered sequences.
Use 'without repetition' when each item can only be selected once (like dealing cards from a deck or choosing team members). Use 'with repetition' when items can be reused (like digits in a PIN code, where 1111 is valid, or selecting ice cream scoops where you can pick the same flavor multiple times).
n represents the total number of distinct items available to choose from. r represents how many items you're selecting or arranging. For example, choosing 5 cards from a 52-card deck means n=52 and r=5. The constraint r ≤ n applies to combinations and permutations without repetition, but r can exceed n when repetition is allowed.
Factorials grow faster than exponential functions. For example: 10! = 3,628,800; 15! = 1,307,674,368,000; 20! ≈ 2.43 × 10^18. This rapid growth is why combinatorial problems quickly produce astronomically large numbers. The calculator handles this by using multiplicative simplification and displays large results in scientific notation.
Stars and bars is a combinatorial technique for distributing r identical items into n distinct bins (or choosing r items from n types with repetition). The formula C(n+r-1, r) counts ways to arrange r stars and n-1 bars. For example, choosing 5 donuts from 3 flavors: n=3, r=5, result = C(3+5-1, 5) = C(7,5) = 21 ways.
Most lotteries use combinations without repetition because order doesn't matter and each number can only appear once. For a 6/49 lottery (pick 6 from 49 numbers), use C(49,6) = 13,983,816. Your odds of winning the jackpot are 1 in ~14 million. For Powerball-style lotteries with a separate bonus ball, multiply the main combination by the bonus ball options.
Permutations count all possible orderings of combinations. The formula P(n,r) = C(n,r) × r! shows that permutations equal combinations multiplied by the number of ways to arrange r items. For example, C(5,3) = 10 (groups of 3 from 5), while P(5,3) = 60 (ordered arrangements of 3 from 5). The ratio is 3! = 6.
By convention, 0! = 1. This isn't arbitrary—it makes the formulas consistent. There's exactly one way to arrange zero items (do nothing), and the combination formula C(n,0) = n!/(0!×n!) should equal 1 (one way to choose nothing). The empty product convention in mathematics also defines an empty product as 1.
For a 4-digit PIN using digits 0-9, use permutations with repetition: n^r = 10^4 = 10,000 possible PINs. Each position can be any of 10 digits, and digits can repeat. For a PIN where digits cannot repeat, use P(10,4) = 10×9×8×7 = 5,040 possible PINs.
Combinations: lottery probability, poker hand odds, selecting committee members, choosing pizza toppings, binomial coefficients in statistics. Permutations: password strength calculation, race finishing orders, seating arrangements, scheduling, encryption key spaces. These concepts are fundamental in probability, statistics, computer science, and cryptography.
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Explore counter-intuitive probability with this famous puzzle
Choose n, r, and whether order or repetition matters to compute the number of possible selections or arrangements.