Combination Calculator

Calculate combinations (nCr) - the number of ways to choose r items from n items when order doesn't matter. Perfect for lottery odds, committee selection, and probability problems.

**Combination Formula:** C(n,r) = n! / (r! � (n-r)!) Also written as: nC? or (n choose r) **Factorial Definition:** n! = n � (n-1) � (n-2) � ... � 2 � 1 0! = 1 (by definition) **Efficient Calculation:** C(n,r) = [n � (n-1) � ... � (n-r+1)] / r! **Important Properties:** � 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) = C(n, n-r) (symmetry property) **Relationship to Permutations:** C(n,r) = P(n,r) / r! P(n,r) = r! � C(n,r) Order doesn't matter in combinations!
**Example 1: Simple Combination** Calculate: C(5,2) - Choose 2 items from 5 Step 1: Apply formula C(n,r) = n! / (r! � (n-r)!) C(5,2) = 5! / (2! � 3!) Step 2: Calculate factorials 5! = 5 � 4 � 3 � 2 � 1 = 120 2! = 2 � 1 = 2 3! = 3 � 2 � 1 = 6 Step 3: Divide C(5,2) = 120 / (2 � 6) = 120 / 12 = 10 **Efficient Method:** C(5,2) = (5 � 4) / (2 � 1) = 20 / 2 = 10 Result: 10 ways to choose 2 items from 5 **Example 2: Committee Selection** Choose 3 people from 10 for a committee Calculate: C(10,3) C(10,3) = 10! / (3! � 7!) = (10 � 9 � 8) / (3 � 2 � 1) = 720 / 6 = 120 Result: 120 different committees possible **Example 3: Lottery** Pick 6 numbers from 49 C(49,6) = 49! / (6! � 43!) = (49 � 48 � 47 � 46 � 45 � 44) / (6 � 5 � 4 � 3 � 2 � 1) = 10,068,347,520 / 720 = 13,983,816 Result: 1 in 13,983,816 chance of winning

What is a combination and how is it different from permutation?

Combinations count selections where ORDER DOESN'T MATTER. Permutations count arrangements where ORDER MATTERS. Example: Choosing 2 letters from {A,B,C}. Combinations: AB, AC, BC (3 ways - AB same as BA). Permutations: AB, BA, AC, CA, BC, CB (6 ways - AB ≠ BA). Formula: C(n,r) = n!/(r!(n-r)!) vs P(n,r) = n!/(n-r)!. Relationship: P(n,r) = r! * C(n,r). Use combinations for: teams, committees, card hands, lottery picks.

What is the combination formula and how do you use it?

Formula: C(n,r) = n!/(r!(n-r)!) = "n choose r". Example: C(5,2) = 5!/(2!*3!) = (5*4*3*2*1)/((2*1)*(3*2*1)) = 120/(2*6) = 10. Simplified method: C(5,2) = (5*4)/(2*1) = 10. Steps: 1) Multiply n down r times (5*4), 2) Divide by r! (2!). Properties: C(n,0)=1, C(n,n)=1, C(n,1)=n, C(n,r)=C(n,n-r). Example: C(10,3) = C(10,7) = 120.

When do you use combinations in real life?

Common applications: 1) Lottery - picking 6 from 49 numbers: C(49,6) = 13,983,816 combinations. 2) Committee selection - choosing 3 people from 10: C(10,3) = 120 ways. 3) Pizza toppings - selecting 3 from 8 toppings: C(8,3) = 56 combos. 4) Poker hands - 5 cards from 52: C(52,5) = 2,598,960. 5) Genetics - allele combinations. 6) Quality control - sample selection. 7) Menu planning - dish combinations. Anytime order doesn't matter in selection.

How do you calculate combinations without a calculator?

Optimization tricks: 1) Use C(n,r) = C(n,n-r), pick smaller: C(20,18) = C(20,2) = 190. 2) Cancel before multiplying: C(10,3) = (10*9*8)/(3*2*1) = (10*9*8)/6. Cancel: 10/2=5, 9/3=3. Result: 5*3*8 = 120. 3) Pascal's Triangle: Each number = sum of two above. Row 5: 1,5,10,10,5,1 gives all C(5,r). 4) Small values memorize: C(n,2) = n(n-1)/2, C(n,3) = n(n-1)(n-2)/6.

What is the relationship between combinations and binomial coefficients?

They're identical! C(n,r) is the binomial coefficient in (a+b)ⁿ expansion. Binomial Theorem: (a+b)ⁿ = Σ C(n,r)aⁿ⁻�^3b�^3. Example: (a+b)^3 = C(3,0)a^3 + C(3,1)a^2b + C(3,2)ab^2 + C(3,3)b^3 = a^3 + 3a^2b + 3ab^2 + b^3. Coefficients (1,3,3,1) are C(3,0) through C(3,3). Pascal's Triangle displays these. Used in: probability distributions (binomial), algebra expansions, probability theory.

Why does C(n,0) equal 1 and what does it mean?

C(n,0) = n!/(0!*n!) = 1 because 0! = 1 by definition. Meaning: there's exactly ONE way to choose nothing from n items - choose nothing! Similarly, C(n,n) = 1 - one way to choose everything. These aren't just math tricks; they're logically consistent. Example: selecting 0 toppings for pizza = 1 way (plain pizza). Empty set is a valid selection. Critical for: binomial theorem (ensures (a+b)⁰=1), probability (P(all fail) calculations), recursion formulas.

How do you solve combination word problems?

Steps: 1) Identify n (total items) and r (items to choose). 2) Verify order doesn't matter. 3) Check for restrictions. Example: "Team of 5 from 12 people, must include Alice" - Fix Alice (she's in), choose 4 from remaining 11: C(11,4) = 330. Multiple groups: "3 men from 7, 2 women from 5" = C(7,3) * C(5,2) = 35 * 10 = 350. Exclusions: "4 people from 10, excluding Bob" = C(9,4) = 126. Practice identifying what's being counted.

What are some important combination properties and identities?

Key properties: 1) Symmetry: C(n,r) = C(n,n-r). 2) Pascal's Identity: C(n,r) = C(n-1,r-1) + C(n-1,r). 3) Sum of row: Σ C(n,r) = 2ⁿ (all subsets). 4) C(n,1) = n, C(n,2) = n(n-1)/2. 5) Vandermonde: C(m+n,r) = Σ C(m,k)*C(n,r-k). 6) Hockey stick: Σ C(r,k) = C(r+1,k+1). Example: C(5,0)+C(5,1)+C(5,2)+C(5,3)+C(5,4)+C(5,5) = 2⁵ = 32 (all possible selections).

How do combinations relate to probability?

Combinations calculate favorable/total outcomes when order doesn't matter. Example: Probability of getting exactly 3 heads in 5 coin flips = C(5,3)/2⁵ = 10/32 = 31.25%. C(5,3) counts arrangements with 3 heads. Poker: P(flush) = (C(13,5)*4)/C(52,5). Lottery: P(winning) = 1/C(49,6) = 1/13,983,816. Binomial probability: P(r successes in n trials) = C(n,r)p�^3(1-p)ⁿ⁻�^3. Hypergeometric: sampling without replacement.

What is the maximum value of C(n,r) for a given n?

Maximum occurs at middle: r = n/2 (if even) or r = (n+/-1)/2 (if odd). Example: For n=6, max at r=3: C(6,3)=20. For n=7, max at r=3 or 4: C(7,3)=C(7,4)=35. This is why: C(n,r) increases as r approaches n/2, then decreases (by symmetry). Pascal's Triangle shows this - middle is largest. Used in: probability (most likely outcome), information theory (maximum entropy), designing experiments. Example: Coin flips - most likely outcome is roughly half heads.