Visual walkthrough — Permutations — nPr, arrangements with restrictions

This page is a slow-motion replay of the parent note's central derivation. If a symbol appears, we define it first.

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Step 1 — What are we even counting?

WHAT. We have a set of distinct objects. "Distinct" just means you can tell them apart — like 5 different letters $A,B,C,D,E$, not 5 identical marbles. We want to line up some of them in a row of positions, and we care about the order.

WHY. Order is the whole point of a permutation. The row $AB$ is a different line-up from $BA$, even though the same two letters were used. If we did not care about order, we'd be selecting a group, which is a different subject (Combinations — nCr).

PICTURE. Look at the figure: the same two tiles placed in two orders give two different arrangements. That "two different pictures from the same tiles" is exactly why order matters.

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Step 2 — The one rule underneath everything: multiply the choices

WHAT. Suppose the first position can be filled in $3$ ways, and for each of those, the second position can be filled in $2$ ways. Then the number of complete rows is $3\times 2=6$.

WHY. This is the Fundamental Counting Principle: when you make one decision and then another, the totals multiply (you don't add them). Think of it as a branching tree — each first choice sprouts a full set of second choices.

PICTURE. The tree below starts with $3$ branches. Each branch splits into $2$. Counting the leaves at the bottom gives $3\times 2=6$ — every leaf is one finished arrangement.

$$\underset{\text{first-slot choices}}{\underbrace{3}}\times \underset{\text{second-slot choices, for each first}}{\underbrace{2}}=\underset{\text{finished rows (leaves)}}{\underbrace{6}}$$

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Step 3 — Watch the choices shrink, position by position

WHAT. Now fill $r$ positions from $n$ distinct objects, no repeats (once a letter is used, it's gone). Fill left to right and record how many are still available at each step.

WHY. "No repeats" is what makes the count drop by exactly one each time. That single shrinking-by-one behaviour is the engine of the whole formula.

PICTURE. The row of boxes below shows the available count above each box: $n$, then $n-1$, then $n-2$, … The pool of leftover tiles on the right physically shrinks by one tile per box.

Position	Available choices	Why one fewer?
1st	$n$	nothing used yet
2nd	$n-1$	one object is now placed
3rd	$n-2$	two are placed
$\dots$	$\dots$	$\dots$
$r$-th	$n-r+1$	$r-1$ already placed

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Step 4 — Multiply the boxes: the raw formula

WHAT. Apply Step 2's multiply-rule across all the boxes from Step 3:

$$n{P}_r=\underset{\text{box 1}}{\underbrace{n}}\cdot \underset{\text{box 2}}{\underbrace{(n-1)}}\cdot \underset{\text{box 3}}{\underbrace{(n-2)}}\cdots \underset{\text{box }r}{\underbrace{(n-r+1)}}$$

WHY. Each box is an independent decision given the ones before it, so by the counting principle their choice-counts multiply. The symbol $n{P}_r$ (read "$n$ permute $r$") is just a name for this product.

PICTURE. The figure stacks the boxes and draws a big $\times$ between them, with the descending numbers $n,n-1,\dots$ written in. It's the product, made visible.

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Step 5 — Compress it with factorials (the clever trick)

WHAT. A factorial $n!$ means "multiply every whole number from $n$ down to $1$": $n!=n(n-1)\cdots 2\cdot 1$. Our raw product stops early — it stops at $n-r+1$ and never reaches the "tail" $(n-r)(n-r-1)\cdots 1$.

WHY. We want a tidy closed form. Trick: multiply and divide by the missing tail $(n-r)!$. Multiplying and dividing by the same thing changes nothing, but it lets the top become a full factorial.

PICTURE. Below, the full ladder $n!$ is drawn top to bottom. The coral part is the piece we actually want ($n{P}_r$); the mint part is the leftover tail $(n-r)!$. Dividing chops the mint tail off.

$$n{P}_r=\underset{\text{what we want (coral)}}{\underbrace{n(n-1)\cdots (n-r+1)}}\cdot \frac{\stackrel{\text{tail (mint)}}{\overbrace{(n-r)!}}}{\underset{\text{same tail, divides out}}{\underbrace{(n-r)!}}}=\frac{n!}{(n-r)!}$$

Let's verify on our running example, $n=5,r=3$:

$$5P_3=\frac{5!}{(5-3)!}=\frac{5!}{2!}=\frac{120}{2}=60,\text{and }5\cdot 4\cdot 3=60.✓$$

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Step 6 — Edge case A: filling every position ($r=n$)

WHAT. What if you arrange all the objects, so $r=n$? The formula gives

$$n{P}_n=\frac{n!}{(n-n)!}=\frac{n!}{0!}.$$

WHY. We need $0!$ to make sense. The rule is $0!=1$ (see Factorials and 0!). With that, $n{P}_n=\frac{n!}{1}=n!$ — arranging all $n$ objects has $n!$ orders, which is exactly what Step 4 predicts (the boxes count down $n,n-1,\dots ,1$).

PICTURE. The figure fills all boxes; the leftover pool becomes empty, and the descending numbers reach all the way down to $1$.

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Step 7 — Edge case B: filling no position ($r=0$)

WHAT. Fill zero positions:

$$n{P}_0=\frac{n!}{(n-0)!}=\frac{n!}{n!}=1.$$

WHY. There is exactly one empty arrangement — the blank row. Not zero (that would mean "impossible"); doing nothing is one valid outcome. The formula agrees: top and bottom are identical, so they cancel to $1$.

PICTURE. An empty row of no boxes, labelled "1 way: the empty line-up." The full pool sits untouched on the side.

Recall Quick self-test

Why is $n{P}_0=1$ and not $0$? ::: Because "arrange nothing" is possible in exactly one way (the empty arrangement); $0$ would mean impossible. What is $7P_7$? ::: $7!=5040$, since $(7-7)!=0!=1$.

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The one-picture summary

Everything above, compressed into a single diagram: row of boxes → shrinking choices → multiply → cancel the unused tail → clean formula. Trace it left to right.

$$\underset{\text{Step 3: choices shrink}}{\underbrace{n,n-1,\dots ,n-r+1}}\stackrel{\times \text{(Step 4)}}{\to }\underset{\text{raw product}}{\underbrace{n(n-1)\cdots (n-r+1)}}\stackrel{\times \frac{(n-r)!}{(n-r)!}\text{(Step 5)}}{\to }\boxed{\frac{n!}{(n-r)!}}$$

Recall Feynman retelling — explain the whole walkthrough to a 12-year-old

Picture lining up friends for a photo but you only have room for a few of them. For the first spot in the row, any friend can stand there — that's a lot of choices, call it $n$. Once one is standing, they're used up, so the next spot has one fewer friend to pick, then one fewer again. To count all possible photos you just multiply "how many are left" at each spot — because each first choice carries a whole fresh set of second choices (that's the branching tree). Multiply the boxes and you get a chain like $5\times 4\times 3$. Mathematicians like tidy formulas, so they notice this chain is just the top part of $n!$ (the full countdown $n$ down to $1$) with the bottom part chopped off. To chop it cleanly they divide by the factorial of "the friends who didn't make it into the photo," $(n-r)!$. That gives the neat $\frac{n!}{(n-r)!}$. And the weird cases behave: photograph everyone and you get $n!$ (because dividing by $0!=1$ does nothing); photograph no one and there's exactly one way — the empty photo.

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Connections

• Fundamental Counting Principle — Step 2's multiply-rule is this principle.
• Factorials and 0! — Steps 5–7 lean on $0!=1$.
• Combinations — nCr — when order stops mattering, divide by $r!$: $n{C}_r=\frac{n{P}_r}{r!}$.
• Circular Permutations — same shrinking idea, bent into a ring.
• Permutations with Repetition — what changes when objects aren't all distinct.
• Probability — Equally Likely Outcomes — these counts become numerators and denominators.