Permutations vs combinations: when order actually matters

Permutations and combinations both count ways to pick r items out of n, but they answer different questions. A permutation counts arrangements, where the order matters. A combination counts selections, where it doesn't. Mixing the two up is the single most common mistake in a probability class, and the fix is one question: if I picked the same items in a different order, is that a different outcome?

The method

Both formulas are built from factorials, where n! means multiply every whole number from n down to 1: 5! = 5 × 4 × 3 × 2 × 1 = 120.

Permutations, order matters: nPr = n! ÷ (n − r)!

Combinations, order doesn't matter: nCr = n! ÷ (r! × (n − r)!)

Notice the second formula is the first one divided by r! — that's not a coincidence, it's the whole relationship: every combination of r items can be arranged in r! different orders, so permutations always outnumber combinations by exactly that factor, for the same n and r.

A real example: a three-person podium (order matters)

Eight runners finish a race; gold, silver and bronze go to the top three. Swapping who's first and who's second genuinely changes the outcome, so this is a permutation of 3 from 8:

8P3 = 8 × 7 × 6 = 336

(That's the full formula 8! ÷ 5! with the common 5! cancelling out, which is why you never need to multiply all the way down to 1 — just the top r numbers.)

The same eight people, no ranking (order doesn't matter)

Now pick 3 of the same 8 runners for an unranked relay team, with no assigned leg order. Swapping two names on the roster doesn't create a new team, so this is a combination:

8C3 = 8P3 ÷ 3! = 336 ÷ 6 = 56

Same 8 people, same group size of 3, and the answer shrank from 336 to 56 — a factor of 6, exactly 3! — because every one of the 56 teams had already been counted 6 times over in the podium version, once for each order its 3 members could stand in.

A real example: choosing a committee

A club of 12 members needs to pick a 4-person committee with no distinct roles — just 4 names, order irrelevant. That's 12C4:

12C4 = (12 × 11 × 10 × 9) ÷ (4 × 3 × 2 × 1) = 11,880 ÷ 24 = 495

495 possible committees from 12 people — a number nobody would want to count by listing them out by hand.

A real example: lottery odds

A classic 6-from-49 lottery draws 6 numbers with no regard to the order they come out in, so the number of possible tickets is 49C6:

49C6 = (49 × 48 × 47 × 46 × 45 × 44) ÷ 720 = 10,068,347,520 ÷ 720 = 13,983,816

Almost 14 million equally likely combinations, which is where the famous "1 in 14 million" lottery-odds figure comes from — it's just 49C6, arrived at with exactly the method above.

Picking the right one

Before reaching for either formula, ask what changes the outcome. Assigning distinct roles, ranks or positions (1st/2nd/3rd, president/treasurer, a PIN code's digit order) means order matters — use permutations. Picking an unranked group (a committee, a hand of cards, a set of lottery numbers) means order doesn't — use combinations. Getting this one question right decides which formula to use before a single number gets multiplied.

Run your own numbers

The free Permutations & Combinations calculator computes both nPr and nCr for any n and r, entirely on your device.