CalcCombinations Calculator
Calculate combinations of n items taken r at a time.
What is a Combination?
A combination counts the number of ways to select r objects from n distinct objects when order does not matter. For example, choosing 3 toppings from 10 options: the selection {pepperoni, mushroom, onion} is identical to {onion, pepperoni, mushroom}. Combinations answer "how many unique groups can I form?"
The combination formula is C(n,r) = n! / (r! × (n-r)!), also written as "n choose r" or ₙCᵣ. This calculates selections where position is irrelevant: lottery numbers, committee members, pizza toppings, or card hands.
Key distinction: Combinations count unordered groups; permutations count ordered arrangements. Choosing 5 cards from a deck uses combinations (the hand is the same regardless of deal order). Arranging those 5 cards in a row uses permutations.
Formulas Explained
Combination formula:
C(n,r) = n! / (r! × (n-r)!)
Where:
n = total number of distinct objects
r = number of objects to select
! = factorial (e.g., 5! = 5×4×3×2×1 = 120)
Alternative notation:
C(n,r) = ₙCᵣ = (n r) = "n choose r" = binomial coefficient
Key properties:
- C(n,r) = C(n, n-r) — choosing r to include equals choosing (n-r) to exclude
- C(n,0) = 1 — one way to choose nothing
- C(n,n) = 1 — one way to choose everything
- C(n,1) = n — n ways to choose one item
- C(n,r) = P(n,r) / r! — combinations = permutations ÷ arrangements
Pascal's Triangle connection:
C(n,r) appears as the r-th entry in row n of Pascal's Triangle. Row sums equal 2^n.
Step-by-Step Guide
- Enter total items (n): The complete set size. Example: 52 cards
- Enter items to select (r): How many to choose. Example: 5 cards
- Click calculate: The calculator computes C(n,r) = n!/(r!×(n-r)!)
- Read your result: For C(52,5): result = 2,598,960 possible hands
- Verify: 52!/(5!×47!) = (52×51×50×49×48)/(5×4×3×2×1) = 2,598,960 ✓
Real Examples with Calculations
Example 1: Poker hand odds
Total 5-card hands from 52-card deck:
C(52,5) = 52! / (5! × 47!) = (52×51×50×49×48) / 120
= 311,875,200 / 120 = 2,598,960 hands
Royal flush: 4 ways (one per suit)
Probability: 4 / 2,598,960 = 1 in 649,740 (0.00015%)
Application: Gambling odds, game design, probability education.
Example 2: Lottery probability
Pick 6 numbers from 1 to 49. How many combinations?
C(49,6) = 49! / (6! × 43!) = (49×48×47×46×45×44) / 720
= 10,068,347,520 / 720 = 13,983,816
Probability of jackpot: 1 in 13.98 million
Application: Lottery design, risk assessment, expected value calculations.
Example 3: Committee formation
Company has 20 employees. How many ways to form a 5-person committee?
C(20,5) = 20! / (5! × 15!) = (20×19×18×17×16) / 120
= 1,860,480 / 120 = 15,504 committees
Application: Team formation, focus groups, board subcommittees.
Example 4: Menu combinations
Restaurant offers 12 side dishes. Choose any 3 for a combo plate.
C(12,3) = 12! / (3! × 9!) = (12×11×10) / 6
= 1,320 / 6 = 220 different combo plates
Application: Menu planning, meal kits, subscription boxes.
Example 5: Clinical trial groups
From 100 patients, select 10 for treatment group.
C(100,10) = 100! / (10! × 90!) = 17,310,309,456,440
That's 17.3 trillion possible treatment groups!
Application: Randomized trials, A/B testing, sampling methodology.
4 Common Mistakes
- Using combinations when order matters: Electing president, VP, and treasurer from 10 people uses permutations P(10,3) = 720, not combinations C(10,3) = 120. Roles are distinct, so order of selection matters.
- Forgetting to divide by r!: C(n,r) = P(n,r) / r!. If you compute n!/(n-r)! without dividing by r!, you get permutations, not combinations. This overcounts by a factor of r!.
- Misapplying the symmetry property: C(n,r) = C(n,n-r) is useful but can confuse. C(10,3) = C(10,7) = 120. Both are correct: choosing 3 to include leaves 7 excluded, and vice versa.
- Assuming all outcomes are equally likely: In poker, C(52,5) gives total hands, but not all hand types have equal frequency. One pair occurs 1,098,240 ways (42% of hands); royal flush occurs 4 ways. Probability requires counting favorable outcomes separately.
4 Pro Tips
- Use symmetry for easier calculation: C(50,47) = C(50,3) = 50×49×48/6 = 19,600. Computing C(50,47) directly requires massive factorials; C(50,3) is trivial.
- Cancel factorials before multiplying: C(52,5) = 52!/(5!×47!) simplifies to (52×51×50×49×48)/(5×4×3×2×1). Cancel 47! from numerator and denominator first.
- Apply the multiplication principle: "Choose 3 men from 8 and 2 women from 6" = C(8,3) × C(6,2) = 56 × 15 = 840. Multiply independent choices.
- Use Pascal's Triangle for small values: Row 5: 1, 5, 10, 10, 5, 1 gives C(5,0) through C(5,5). Quick reference without calculation. Each entry is sum of two entries above it.
FAQs
Combinations ignore order, so each group is counted once. Permutations count all r! orderings of each group. For C(10,3) = 120 groups, each group has 3! = 6 orderings, giving P(10,3) = 120 × 6 = 720.
C(n,0) = 1 for any n. There is exactly one way to choose zero items: choose nothing. The empty set is unique. Mathematically: n!/(0!×n!) = n!/(1×n!) = 1.
The binomial theorem uses C(n,r) as coefficients: (a+b)^n = Σ C(n,r) × a^(n-r) × b^r. For (x+1)^4: coefficients are C(4,0)=1, C(4,1)=4, C(4,2)=6, C(4,3)=4, C(4,4)=1, giving x^4 + 4x³ + 6x² + 4x + 1.
Yes, when outcomes are equally likely. Probability = C(favorable) / C(total). Example: Probability of exactly 3 aces in 5-card hand = C(4,3)×C(48,2) / C(52,5) = 4×1,128 / 2,598,960 = 0.17%.
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