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

A permutation is an ordered arrangement of items. Order matters — ABC and BAC are different permutations. A combination is an unordered selection — {A,B,C} and {B,A,C} are the same combination. Example: choosing 2 letters from A, B, C. - Permutations P(3,2): AB, BA, AC, CA, BC, CB = 6 - Combinations C(3,2): {A,B}, {A,C}, {B,C} = 3 - Ratio: P(3,2)/C(3,2) = 6/3 = 2 = 2! (arrangements of 2 items) When to use permutations: - Race finishing order (1st, 2nd, 3rd from 10 runners) - PIN codes and passwords (order matters) - Seating arrangements - Any problem where position or sequence matters When to use combinations: - Choosing a committee from candidates - Selecting lottery numbers - Picking pizza toppings - Any problem where only membership matters

How do you calculate P(n,r) step by step?

P(n,r) = n! / (n-r)! which simplifies to a descending product of r terms starting at n. For P(10,3): Step 1: n=10, r=3 Step 2: P(10,3) = 10! / 7! = (10 x 9 x 8 x 7!) / 7! Step 3: Cancel 7!: = 10 x 9 x 8 Step 4: = 720 For P(8,4): 8 x 7 x 6 x 5 = 1,680 For P(52,3) — ordered 3-card hands: 52 x 51 x 50 = 132,600 Key shortcut: always write the product as r consecutive integers counting down from n. No need to compute large factorials explicitly — the (n-r)! cancels completely.

What are real-world examples of permutations?

Permutations arise whenever sequence or position matters. Sports and rankings: - Olympic medal positions (gold/silver/bronze) from 8 finalists: P(8,3) = 336 - Horse racing trifecta (1st, 2nd, 3rd in order) from 12 horses: P(12,3) = 1,320 Codes and passwords: - 4-digit PIN from digits 0-9 (no repeats): P(10,4) = 5,040 - 4-letter code from 26 letters (no repeats): P(26,4) = 358,800 - Lock combination (though lock combinations are actually permutations, not combinations) Seating and arrangement: - 5 people in 5 chairs: P(5,5) = 5! = 120 - Arranging 3 books from a shelf of 10: P(10,3) = 720 Cards: - Ordered 3-card hand from 52: P(52,3) = 132,600 - Compare to unordered C(52,3) = 22,100 — permutations are 6x more (= 3!)

What is the relationship between P(n,r) and n factorial?

P(n,n) = n! — arranging all n items gives n factorial total arrangements. P(n,r) is a partial factorial: it multiplies only the top r terms of n!. Factorial values: - 0! = 1 - 1! = 1 - 5! = 120 - 10! = 3,628,800 - 20! = 2,432,902,008,176,640,000 - 52! is a 68-digit number (more than the estimated atoms in the observable universe is 10^80) P(n,r) avoids computing the full n! by cancelling the (n-r)! denominator: P(52,5) = 52!/47! = 52 x 51 x 50 x 49 x 48 = 311,875,200 This calculator uses BigInt arithmetic for exact results — regular JavaScript numbers lose precision above 2^53, so C(100,50) and P(100,50) would be wrong without BigInt.

How does P(n,r) grow with n and r?

Permutations grow very quickly with both n and r. Fixing r=3, varying n: - P(5,3) = 60 - P(10,3) = 720 - P(20,3) = 6,840 - P(50,3) = 117,600 - P(100,3) = 970,200 Fixing n=10, varying r: - P(10,1) = 10 - P(10,2) = 90 - P(10,3) = 720 - P(10,5) = 30,240 - P(10,10) = 3,628,800 = 10! Growth rate: P(n,r) grows polynomially in n for fixed r (degree r polynomial) and exponentially in r for fixed n. The jump from r=9 to r=10 when n=10 multiplies by 10/1 = 10, while n=100 to n=101 at r=3 multiplies by approximately (101/98)^3.

Permutation Calculator (nPr) — P(n,r) = n! / (n-r)!

The Permutation Calculator computes P(n,r) — the number of ordered arrangements of r items chosen from n items. Uses exact BigInt arithmetic for precise results on large values. Shows the expanding product n x (n-1) x ... x (n-r+1), the step-by-step factorial cancellation, a side-by-side comparison ...

P(n, r) = ?

ARRANGE r ITEMS FROM n ITEMS (ORDER MATTERS)

n (total)

r (arrange)

)

P(10, 3) = ORDERED ARRANGEMENTS

720

= 10! / 7! = 10 × 9 × 8

Permutations P(10,3)

720

order matters — this result

Combinations C(10,3)

120

order ignored

Ratio P / C = r!

3! arrangements per group

Probability (1/nPr)

0.00138889

of one specific arrangement

STEP-BY-STEP

n = 10, r = 3

P(n,r) = n! / (n−r)!

P(10,3) = 10! / 7!

= 10 × 9 × 8 = 720

Comparison: C(10,3) = 120

P(10,3) / C(10,3) = 6 = 3!

P(n,0) = 11

P(n,1) = n10

P(n,n) = n!3,628,800

P(n,2)90

QUICK EXAMPLES

ARRANGEMENT DIAGRAM

Choices available for each position slot

PERMUTATION vs COMBINATION

Property	Permutation P(n,r)	Combination C(n,r)
Order matters	Yes	No
Formula	n!/(n-r)!	n!/(r!(n-r)!)
Result	720	120
Ratio P/C	3! = 6	1
Use case	Rankings, codes, order	Groups, teams, selection

WHAT THIS MEANS

There are 720 ways to arrange 3 items chosen from 10 in order.

Created with❤️byeaglecalculator.com

FORMULAS

PERMUTATION & COMBINATION FORMULAS

NAME	FORMULA	DESCRIPTION
Permutation nPr	P(n,r) = n! / (n-r)!	Ordered arrangements of r items from n
Expanded product	P(n,r) = n x (n-1) x ... x (n-r+1)	Multiply r consecutive integers down from n
Combination nCr	C(n,r) = n! / (r! x (n-r)!)	Unordered selections — order does not matter
Relationship	P(n,r) = C(n,r) x r!	Permutations = combinations x arrangements of r chosen items
Special P(n,n)	P(n,n) = n!	All n items arranged: factorial of n
Special P(n,0)	P(n,0) = 1	Arranging zero items has exactly one way
Special P(n,1)	P(n,1) = n	Arranging one item has n choices
Probability	p = 1 / P(n,r)	Probability of guessing one specific arrangement

HOW TO USE

1
Enter n — the total number of items. For example, n=26 for the alphabet or n=52 for a deck of cards.
2
Enter r — how many items you are arranging. Order matters: ABC and BAC are two different permutations.
3
P(n,r) appears instantly as an exact integer. The arrangement diagram shows the number of choices for each position slot.
4
Compare with C(n,r) in the results — permutations always equal or exceed combinations. The ratio P(n,r)/C(n,r) = r!, the number of ways to arrange the chosen items.
5
Check special cases: P(n,0)=1 (one way to arrange nothing), P(n,1)=n, P(n,n)=n! (all items in every order).

WORKED EXAMPLE

P(10,3): n=10, r=3. P(10,3)=10!/7!=10x9x8=720. C(10,3)=120. Ratio 720/120=6=3!. Probability=1/720=0.00139. P(52,3)=52x51x50=132,600 ordered 3-card hands. P(5,5)=5!=120 (all 5 items). P(26,4)=26x25x24x23=358,800 letter codes.

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Last updated: April 29, 2026 · Formula verified by EagleCalculator team · Eagle-eyed accuracy for every calculation.