Permutation Calculator

Calculate permutations (nPr) - the number of ways to arrange r items from n items when order matters. Supports standard, circular, and repetition permutations.

**Standard Permutation:** P(n,r) = n! / (n-r)! = n � (n-1) � (n-2) � ... � (n-r+1) **All Permutations:** P(n) = n! = n � (n-1) � (n-2) � ... � 2 � 1 **Permutation with Repetition:** P_rep(n,r) = n? Each position has n independent choices **Circular Permutation:** P_circular(n) = (n-1)! Fix one position, arrange the rest **Permutation with Identical Items:** P = n! / (n1! � n2! � ... � n?!) where n? is the count of each identical item type **Relationship to Combinations:** P(n,r) = r! � C(n,r) C(n,r) = P(n,r) / r! Order matters in permutations!
**Example 1: Standard Permutation** Calculate: P(5,3) - Arrange 3 items from 5 Step 1: Apply formula P(n,r) = n! / (n-r)! P(5,3) = 5! / 2! Step 2: Calculate factorials 5! = 5 � 4 � 3 � 2 � 1 = 120 2! = 2 � 1 = 2 Step 3: Divide P(5,3) = 120 / 2 = 60 **Efficient Method:** P(5,3) = 5 � 4 � 3 = 60 Result: 60 different arrangements **Example 2: Race Podium** Top 3 positions from 10 runners Calculate: P(10,3) P(10,3) = 10! / 7! = 10 � 9 � 8 = 720 Result: 720 possible podium arrangements (1st, 2nd, 3rd matter) **Example 3: All Arrangements** Arrange all 4 books on a shelf Calculate: P(4) = 4! 4! = 4 � 3 � 2 � 1 = 24 Result: 24 different arrangements **Example 4: With Repetition** 4-digit PIN code (digits 0-9 can repeat) Calculate: P_rep(10,4) = 104 104 = 10 � 10 � 10 � 10 = 10,000 Result: 10,000 possible PIN codes **Example 5: Circular Arrangement** Seat 6 people around a round table Calculate: P_circular(6) = (6-1)! 5! = 5 � 4 � 3 � 2 � 1 = 120 Result: 120 arrangements (vs 720 for a line)

What is a permutation and when do you use it?

Permutations count ARRANGEMENTS where ORDER MATTERS. Use when sequence/position is important. Example: Password "ABC" ≠ "BAC". 3 letters from {A,B,C}: ABC, ACB, BAC, BCA, CAB, CBA = 6 permutations. Formula: P(n,r) = n!/(n-r)!. Example: P(5,2) = 5!/3! = 5*4 = 20. Use for: passwords, race rankings, seating arrangements, lock combinations (actually permutations!), order of tasks, arranging books on shelf.

What is the difference between permutations and combinations?

PERMUTATIONS: Order matters. COMBINATIONS: Order doesn't matter. Example: 2 from {A,B,C}. Permutations (6): AB, BA, AC, CA, BC, CB. Combinations (3): AB, AC, BC (AB=BA counted once). Formula relationship: P(n,r) = r! * C(n,r). Example: P(5,3) = 60 = 6 * C(5,3) = 6 * 10. Mnemonic: Permutation has "Position", Combination has "Choose". Lock = permutation (1-2-3 ≠ 3-2-1). Lottery = combination (numbers in any order).

What is the permutation formula and how do you calculate it?

Standard: P(n,r) = n!/(n-r)! = n(n-1)(n-2)...(n-r+1) for r items. Example: P(7,3) = 7!/4! = 7*6*5 = 210. All items: P(n) = n!. Example: arranging 5 books = 5! = 120. With repetition: n�^3. Example: 3-digit code from 10 digits = 10^3 = 1000. Circular: (n-1)!. Example: 6 people around table = 5! = 120. Key insight: Each position has fewer choices as items are used.

What are permutations with repetition and when do you use them?

When items can be reused, use n�^3 (not n!/(n-r)!). Each position has n choices independently. Example: 4-digit PIN from 10 digits = 10⁴ = 10,000 (can repeat digits like 1111). Compare without repetition: P(10,4) = 5,040. Applications: passwords allowing repeats, dice rolls (each die independent), license plates with repeated letters, binary strings (2ⁿ), sampling with replacement. Formula simpler but gives more possibilities.

What are circular permutations and how do they differ?

For arranging items in a circle, rotations are considered identical. Formula: (n-1)! instead of n!. Why? Fix one position as reference (rotation doesn't create new arrangement). Example: 4 people at round table = 3! = 6 (vs 4!=24 in line). Visual: ABCD, BCDA, CDAB, DABC are same circle. If reflections also identical (like necklace flipped): (n-1)!/2. Applications: seating at round tables, arranging beads on bracelet, people in circle game, molecular structures.

How do you handle permutations with identical items?

Formula: n! / (n₁! * n₂! * ... * nₖ!) where nᵢ = count of identical items. Example: Letters in "MISSISSIPPI" (11 letters: 1M, 4I, 4S, 2P) = 11!/(1!*4!*4!*2!) = 34,650. Without division: overcounts identical arrangements. Example: "AAB" = 3!/2! = 3 (AAB, ABA, BAA not 6). Applications: arranging books with duplicates, word anagrams, arranging colored balls, distinguishable vs indistinguishable objects.

What are common real-world applications of permutations?

Everyday uses: 1) Passwords/PINs - 8 chars from 26 letters = P(26,8) or 26⁸ with repetition. 2) Race results - 1st, 2nd, 3rd from 10 runners = P(10,3) = 720. 3) Seating arrangements - 6 people in 6 chairs = 6! = 720. 4) Playlist order - 20 songs shuffled = 20! ≈ 2.4*10^1⁸. 5) Task scheduling - order matters. 6) Travel routes - visiting cities in sequence. 7) Genetic sequences - DNA base arrangements. 8) Tournament brackets - game order.

How do you solve permutation word problems?

Steps: 1) Identify if order matters (yes = permutation). 2) Find n (total) and r (selecting). 3) Check restrictions. Example: "Arrange 3 of 5 books" = P(5,3) = 60. Restrictions: "5 people, Alice must be first" = fix Alice, arrange 4 = 4! = 24. "4 digits, first can't be 0" = 9*10*10*10 = 9,000. Multiple groups: "President, VP from 10 people" = P(10,2) = 90. Adjacent items: "Keep 2 together" = treat as unit, (n-1)! * 2!.

What is the relationship between factorials and permutations?

Factorial is special case of permutation: n! = P(n,n) = arranging all n items. P(n,r) = n!/(n-r)! uses factorial ratio. Example: P(6,4) = 6!/2! = 720/2 = 360. Factorials grow extremely fast: 10! = 3,628,800; 20! ≈ 2.4*10^1⁸; 70! > atoms in universe. Computational challenge for large n. Stirling's approximation: n! ≈ sqrt(2PIn)(n/e)ⁿ for estimates. Calculator limits around 170! (overflow). Use logarithms for comparisons.

Can you have more arrangements than combinations, and by how much?

Always P(n,r) >= C(n,r). Relationship: P(n,r) = r! * C(n,r). The difference is r! factor (ordering r items). Example: P(10,4) = 5,040 vs C(10,4) = 210. Ratio = 4! = 24 (each combination creates 24 permutations). Larger r = bigger ratio. At r=n: P(n,n) = n! vs C(n,n) = 1. Extreme: P(10,10) = 3,628,800 vs C(10,10) = 1 (ratio = 10!). For small r: P(10,2) = 90 vs C(10,2) = 45 (ratio = 2!).