When order counts too: line them up and multiply the shrinking choices.
In the last lesson you learned to multiply the steps of a job and skip writing the list. Here we put that to work on a question that shows up everywhere: out of n things, choose m of them and line them up in order — how many arrangements are there? When order matters, AB and BA are two different answers, not one. We call an ordered selection a permutation, and counting them turns out to be nothing more than a relay of shrinking choices: the first slot has n options, the next has n−1, and so on. Multiply those and you have the whole story — a tidy formula, a name for "multiply all the way down" (the factorial), and a handful of tactics for the puzzles where someone must go first, or two people must sit together, or a number can't start with zero.
Three runners — A, B, C — finish a race. The judge reads off first, second, third. Is "A, B, C" the same result as "B, A, C"? Of course not — the gold medal changed hands. When the order of a selection carries meaning, each ordering is its own outcome.
A permutation is an ordered selection: you pick some objects and arrange them in a sequence. Because order counts, AB ≠ BA — swapping two items gives a genuinely different permutation. We write the number of ways to fill m ordered places from n distinct objects as P(n, m).
Lining up all three runners gives six finishing orders: ABC, ACB, BAC, BCA, CAB, CBA. Six — not three, because the order is part of the answer. Below is every one of them; notice each is a full reordering of the same three letters.
Now the engine. To fill m ordered places from n objects, think of the places as a row of slots and fill them left to right, one decision at a time — exactly the multiplication principle from 32.1.
The first slot can take any of the n objects. Once it is filled, one object is used up, so the second slot has only n−1 left. The third has n−2, and so on. After filling m slots you have peeled off m factors, each one smaller than the last:
P(n, m) = n · (n−1) · (n−2) · … · (n−m+1)
That last factor deserves a glance: after using m−1 objects, the count is n−(m−1) = n−m+1. So a permutation count is always m shrinking factors, the largest being n. For three places out of five: P(5, 3) = 5 · 4 · 3 = 60 — three factors, starting at five.
A club of 10 people elects a president and a vice-president — two ordered jobs, since the roles differ. P(10, 2) = 10 · 9 = 90. Ten choices for president, nine left for vice-president; ninety ticket combinations.
What if you line up all n objects, leaving none behind? Then m = n, and the shrinking factors run all the way down to 1:
P(n, n) = n · (n−1) · … · 2 · 1 = n!
That product has a name: n factorial, written n! with a real exclamation mark. So P(4, 4) = 4! = 24, and P(3, 3) = 3! = 6 — exactly the six orderings of A, B, C we listed. Factorials grow ferociously: 5! = 120, 10! = 3,628,800.
By agreement, 0! = 1. It looks odd, but it makes the formulas seamless: there is exactly one way to arrange nothing (do nothing), and the rule below works without exception only if 0! = 1.
Now compare the partial product with the full one. The shrinking factors of P(n, m) are the top of n! — they stop at n−m+1, and what gets chopped off is exactly (n−m)!. So multiply and divide by that leftover and the whole thing collapses into one clean closed form:
P(n, m) = n(n−1)…(n−m+1) = n!(n−m)!
Check it on P(5, 3): 5!2! = 120/2 = 60. Same sixty as the slot product — the dividing-out of (n−m)! just cancels the factors you never used. (On a calculator this is the nPr button; some textbooks write it A(n, m) or Anm — same number.)
Real problems add conditions: "the captain must lead," "she can't go last," "this seat is reserved." The master tactic is always the same — handle the fussy position or person first, while it still has the fewest options, then arrange whatever is left with an ordinary permutation.
Five students line up for a photo, and A insists on standing first. Settle A's spot first: only 1 way. The other four fill the remaining four places freely: P(4, 4) = 4! = 24. Total = 1 × 24 = 24.
Same five students, but now A refuses to be last. The cleanest route is total minus the bad ones. All line-ups: 5! = 120. The forbidden ones have A last — fix that spot, arrange the other four: 4! = 24. So the answer is 120 − 24 = 96.
Both moves share one instinct: deal with the restriction before it can trip you. Pin down the special spot, or subtract the line-ups that break the rule — whichever is simpler to count.
Two conditions come up so often they are worth a tactic each.
If two people must stand side by side, glue them into a single block and arrange the blocks. Then remember the pair can face two ways inside the block, so multiply by the block's internal arrangements.
Arrange A, B, C, D with C and D together. Treat [CD] as one object: now you are arranging three things — A, B, [CD] — in 3! = 6 ways. Inside the block, CD or DC: 2! = 2 ways. Multiply: 6 × 2 = 12.
If two people must not touch, seat everyone else first, which opens up gaps between and around them. Then drop the two into separate gaps — and since the two people are distinct and the gaps are different positions, order matters, so use a permutation of the gaps.
Same four letters, now C and D apart. Seat A and B: 2! = 2 ways. That makes 3 gaps — _ A _ B _ — and you place C, D into two different gaps in order: P(3, 2) = 3 · 2 = 6. Multiply: 2 × 6 = 12.
Every arrangement of four letters has C, D either together or apart — never both, never neither. So together + apart should equal all the line-ups: 12 + 12 = 24 = 4!. It checks. ✓
The richest permutation puzzles stack a few conditions, and the classic is making numbers from digits. Each digit-place is a slot, so it is a permutation — but numbers carry a hidden rule that letters do not: a number cannot start with 0. The trap is to forget it, count 0 as a legal leading digit, and overcount.
"032" is not a real three-digit number — it's just 32 with a stray zero. So in a digits-to-number problem, the first slot is special: handle it before the others, and don't let 0 sit there.
How many three-digit numbers, no repeated digits, using digits from {1, 2, 3, 4, 5}? No zero anywhere, so nothing special about the first slot — it's a plain permutation: P(5, 3) = 5 · 4 · 3 = 60.
Now use digits {0, 1, 2, 3, 4}, three-digit numbers, no repeats. Settle the leading slot first. It can't be 0, so it has 4 choices (1–4). The remaining two slots draw from the four digits still unused — including 0 now — in order: P(4, 2) = 4 · 3 = 12. Multiply: 4 × 12 = 48.
Cross-check with total minus bad: pretend 0 were allowed up front, P(5, 3) = 60; subtract the ones that start with 0, which fix 0 in front and arrange two of the remaining four, P(4, 2) = 12. So 60 − 12 = 48. Same answer. ✓
That is the whole craft of combined problems: read each condition, decide which slot it constrains, and fill that slot first. Once the special place is settled, the rest is an ordinary P(n, m).
| Idea | Formula / tactic | Quick example |
|---|---|---|
| Ordered pick of m from n | P(n, m) = n(n−1)…(n−m+1) | P(5, 3) = 5·4·3 = 60 |
| Closed form | P(n, m) = n! / (n−m)! | 5!/2! = 120/2 = 60 |
| Full permutation | P(n, n) = n!, 0! = 1 | P(4, 4) = 4! = 24 |
| Special spot / person | settle it first, then P(rest) | A first: 1 × 4! = 24 |
| Must be together | bundle as a block × inside | 3! × 2! = 12 |
| Must be apart | arrange rest, slot into gaps | 2! × P(3, 2) = 12 |
| Digits → number | first slot ≠ 0, fill it first | {0–4}: 4 × P(4, 2) = 48 |
One sentence to keep: a permutation is an ordered pick, and its count is a product of shrinking choices, P(n, m) = n!/(n−m)!. Strip the order back out — when AB and BA count as the same choice — and you get a combination, the subject of the next lesson. (Look back at the two counting principles if "multiply the steps" still feels new.)
Compute P(7, 3) as a product of shrinking factors, then as a factorial fraction.
In how many ways can 6 distinct books be arranged in a row on a shelf?
Five people sit in a row. A must sit at one of the two ends. How many seatings?
Four friends A, B, C, D line up, but A and B insist on standing next to each other. How many line-ups?
How many three-digit numbers with no repeated digit can be made from {0, 1, 2, 3, 4, 5}?
In how many ways can 5 students stand in a row so that A and B are not next to each other?
Six questions to lock it in. Tap the answer you think is right.
This lesson develops permutations toward CCSS HSS-CP.B.9 — use permutations and combinations to compute probabilities of compound events. The throughline is that an ordered selection is just the multiplication principle applied to slots with shrinking choice-counts, giving P(n, m) = n(n−1)…(n−m+1) = n!/(n−m)!. Encourage learners to verbalize why each factor shrinks rather than memorizing the formula, and to name the tactic before computing on restricted problems (settle the special slot first, bundle for "together," gaps for "apart," and guard the leading zero in digit problems). The "together + apart = all arrangements" check is a gentle, self-correcting habit worth reinforcing.