What is the difference between a permutation and a combination?
Permutations count arrangements where order matters; combinations count selections where it doesn't. Picking a president and vice-president from 10 people: 10P2 = 90 (order matters). Picking 2 committee members: 10C2 = 45 (order doesn't). Every combination corresponds to r! permutations, which is why nPr = nCr × r!.
How are nPr and nCr calculated?
nPr = n!/(n−r)! — multiply n × (n−1) × … for r factors. nCr = n!/(r!(n−r)!) — the same, divided by r! to cancel the orderings. Example: 10P3 = 10×9×8 = 720; 10C3 = 720/6 = 120. This calculator computes both exactly using big-integer arithmetic, so even large results are precise.
When do I use each in probability problems?
Use combinations for lottery-style problems ("chance of matching 6 of 49 numbers" → 49C6), card hands, and group selection. Use permutations when positions are distinct — race finishing orders, PINs without repeated digits, seating arrangements. The classic test: if swapping two chosen items creates a "different" outcome, order matters — use permutations.
What are 0! and nC0?
By definition 0! = 1, which makes the formulas consistent: nC0 = 1 (one way to choose nothing) and nCn = 1 (one way to choose everything). Similarly nP0 = 1. These edge cases aren't arbitrary — they're required for the binomial theorem and recursive identities to hold.
What if some of the items are identical?
Then the plain factorial overcounts, because swapping two identical letters produces an arrangement you cannot tell apart from the original. Divide by the factorial of each repeat count to correct it. The word MISSISSIPPI has eleven letters with four S, four I, two P and one M, so the number of distinguishable arrangements is 11! divided by (4! times 4! times 2!), which comes to 34,650 rather than the 39,916,800 that 11! alone would suggest. The same correction applies to arranging coloured beads, scheduling identical shifts, or counting routes on a grid where the moves in each direction are interchangeable.
How do I count arrangements around a circle?
Fix one position and arrange the rest, giving (n-1)! rather than n!. The reason is that a circular seating plan has no first chair — rotating everybody one place to the left produces the same arrangement of neighbours, so all n rotations of a layout are really one layout. Eight people around a table therefore have 7! = 5,040 distinct arrangements, not 40,320. If the arrangement can also be flipped over and still count as the same, as with a bead necklace or a bracelet, divide again by two to get (n-1)!/2. Deciding which of the three counts applies is usually harder than the arithmetic.