We learn to count as toddlers — one, two, three, point at each duck. That kind of counting works great until the things you want to count won’t sit still long enough to point at them. How many different passwords? How many ways to seat eight people at dinner? How many lottery tickets would I need to guarantee a win?

You can’t point your way through those. But you can still get an exact answer — often a comically large one — without listing a single case. That’s the whole trick of combinatorics: counting without enumerating. This post builds it up from scratch. There’s a widget every few paragraphs; the math will stick better if you actually play with them.

The one idea everything rests on

Forget formulas for a second. Here’s the only principle you really need:

If one choice has A options and a second, independent choice has B options, then making both choices has A × B outcomes.

Multiply, don’t add. That’s it. Adding is for “how many ducks total”; multiplying is for “how many combinations of choices.” It feels almost too simple, so let’s watch it happen. Build a sundae: pick how many scoop flavors, sauces, and toppings the shop offers, and count the distinct sundaes.

Drag a slider and the tree fans out. Every scoop sprouts a full copy of the sauce choices; every sauce sprouts a full copy of the toppings. Three scoops, two sauces, three toppings isn’t 3 + 2 + 3 = 8 sundaes — it’s 3 × 2 × 3 = 18, because each earlier choice gets to pair with all of the later ones.

This scales terrifyingly well, which is the point. A 4-digit PIN is four independent choices, each with 10 options: 10 × 10 × 10 × 10 = 10,000. A 6-character password using lowercase letters and digits is 36⁶ ≈ 2.2 billion. Nobody enumerated those. We just multiplied.

When the choices use each other up

The sundae choices were independent — picking chocolate sauce didn’t remove a scoop flavor from the menu. But a lot of counting problems have choices that consume the options.

Say eight runners line up for a race. How many ways can the podium — gold, silver, bronze — turn out? Gold has 8 possible runners. But once gold is decided, only 7 runners are left for silver. Then 6 for bronze. So it’s 8 × 7 × 6 = 336. Still the multiplication principle, just with a shrinking menu.

When you keep multiplying a shrinking count all the way down to 1 — 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 — you get a factorial, written 8!. That’s the number of ways to arrange all eight runners in order: 40,320. Factorials are what you get when order matters and you use up everything.

Here’s a lab for arranging things. Pick a few items and how many slots to fill, then click tiles to build orderings. Each distinct ordering you complete gets recorded. Try to find them all — and notice how fast “all” gets out of hand.

An ordered arrangement like this is a permutation. When you fill every slot (r = n), you’re counting n! total arrangements. When you fill fewer slots than you have items (r < n), you’ve got a partial permutation — the race podium, where you arrange 3 of 8 — and the count is

That formula is less scary than it looks. It just says: start the factorial n × (n−1) × …, but stop after r factors. Dividing by (n − r)! is the bookkeeping that chops off the tail you didn’t use. For the podium: 8! / 5! = 8 × 7 × 6 = 336. Same answer as before.

The plot twist: sometimes order is a lie

Everything so far assumed order matters — that gold-then-silver is a different outcome from silver-then-gold. For a race, sure. But swap the race for a different question:

Eight runners, but now I’m just picking 3 of them for a relay team — no positions, just “you three.”

Suddenly {Ann, Bo, Cy} and {Cy, Ann, Bo} are the same team. The permutation machine counted them as different, and it counted every other shuffle of those three names too. It overcounted, and it overcounted by an exact, predictable amount.

This is the single most important idea in the whole post, so here’s a widget for it. Each row is one underlying group; the cells in a row are all the different orderings of that same group. Flip “order matters” off and watch each row collapse.

When order is on, you’re looking at nPr — every ordering is its own outcome. Flip it off and each row’s orderings crunch down into a single survivor. The number of orderings that collapse per row is always the same: it’s r!, the number of ways to shuffle r chosen items among themselves. Three people can be ordered 3! = 6 ways, so the permutation count over-counted each team exactly 6-fold.

So to count unordered selections — combinations — you take the permutation count and divide out the redundancy:

That’s the famous “n choose r.” It’s just permutations with the order-double-counting refunded.

Both formulas, side by side

Here’s the calculator. Slide n and r and watch the two numbers diverge — and notice that the gap between them is exactly r!. The Pascal’s triangle at the bottom is a bonus: every nCr you can ask for is already sitting in it, and each number is just the sum of the two above it.

A few things worth pausing on as you slide:

• r = 0 and r = n both give 1 combination. There’s one way to pick nobody, and one way to pick everybody. The formula handles it because 0! = 1 — a convention that exists precisely so this comes out right.
• nCr is symmetric: choosing 3 of 12 to include is the same as choosing 9 of 12 to leave out, so 12C3 = 12C9. The triangle is a mirror.
• Permutations explode faster than combinations, and the ratio between them is always r!. By the time you’re choosing 6 of 12, order-matters gives 665,280 and order-ignored gives 924 — a 720× gap, which is exactly 6!.

So, the lottery

Back to the question at the top. A typical lottery: pick 6 numbers out of 49, order irrelevant. That’s a combination — 49C6. Plug it in (or trust the calculator): 13,983,816. Just shy of 14 million.

If order had mattered, it’d be 49P6 ≈ 10 billion. The lottery is “only” 14 million precisely because they refund the 6! = 720 ways to reorder your picks. Cold comfort: buying one ticket a week, you’d expect to wait about 270,000 years for a jackpot. Buying every combination to guarantee a win would cost ~$28 million in $2 tickets — occasionally a real arbitrage when jackpots roll over, and a few syndicates have actually done it.

The whole toolkit on one card

That’s genuinely most of introductory combinatorics. The decision tree is small:

• Do the choices multiply independently? → multiplication principle, A × B × C … (sundaes, PINs, passwords).
• Are you arranging things in order, using them up? → permutations, nPr = n! / (n−r)! (race podiums, seating charts, rankings).
• Are you just selecting a group, order be damned? → combinations, nCr = n! / (r!·(n−r)!) (teams, committees, lottery tickets, poker hands).

Ninety percent of the work in any real problem is just figuring out which of those three you’re looking at — and the tell is almost always the same one question: does reordering my answer make it a different answer? If yes, you’re permuting. If no, divide by r! and you’re combining.

The duck-pointing kind of counting tops out around the number of ducks you can stand to look at. This kind doesn’t top out at all. That’s the upgrade.