Calculate permutations and combinations (nPr, nCr) with step-by-step factorial breakdown.
Given a set size n and a selection size r, this calculator returns the two counting numbers built on factorials: nPr, the ordered ways to choose r items, and nCr, the ways when order does not matter. The factorial expansion is written out for each, so the arithmetic can be checked line by line. The mistake most people make is computing the wrong one of the two: whether order matters is the entire difference, and nPr is always nCr times r!. Scale is the other surprise - 8P3 is 336, while 20P10 is past 670 billion, so a result that looks too large is often exactly right.
A concrete input and expected output from the current implementation.
Input
n = 8, r = 3
Expected output
nPr (permutations): 8P3 = 8! / (8 - 3)! = 40320 / 120 = 336 nCr (combinations): 8C3 = 8! / (3! x (8 - 3)!) = 40320 / (6 x 120) = 40320 / 720 = 56
Ordered picks fill 3 slots with 8, then 7, then 6 choices, giving 8 x 7 x 6 = 336. Combinations count the 3! = 6 orderings of the chosen trio as one result, so 336 / 6 = 56. The pair always differs by exactly r!.
Both answers come from one pattern. nPr = n! / (n - r)! counts ordered arrangements: fill r slots one at a time, with one fewer choice per slot, which equals n! divided by the leftover (n - r)!. nCr = n! / (r! x (n - r)!) takes that same quotient and divides by r! again, stripping out the r! internal orderings of the chosen items. The calculator substitutes the factorial values and performs the division.
The step-by-step view expands every factorial into its factors before dividing. For 8P3 it writes 8! as 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1, cancels the shared 5 x 4 x 3 x 2 x 1 against the denominator, and is left with 8 x 7 x 6 = 336. For 8C3 it leaves 8 x 7 x 6 over 3 x 2 x 1, or 336 / 6 = 56. Every line restates a step you could redo by hand.
Every set of r chosen items can be arranged in r! different orders, and nPr counts each arrangement separately, while nCr counts the set once. So nPr = nCr x r! exactly, every time: for n = 8, r = 3, that is 56 x 6 = 336. The gap widens fast because r! grows faster than any fixed multiplier.
You get the denominator, which is the hard part. A game drawing 6 balls from 49 has 49C6 = 13,983,816 possible tickets, so one entry wins with probability 1 in 13,983,816. Converting that count into odds or a percentage is on you, and you must check the game rules first: with replacement, or with order counted, the formula changes.
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