Permutations — nPr, arrangements with restrictions: Worked examples

Before we start, one reminder of the ONE tool underneath everything: the multiplication principle from Fundamental Counting Principle — "if a first choice can be made in $a$ ways and a following choice in $b$ ways, the pair can be made in $a \times b$ ways." Every count below is just this, applied slot by slot.

The scenario matrix

Think of each permutation problem as landing in exactly one cell of this table. The rightmost column names the worked example (E1–E9) that lands there.

#	Case class	What makes it tricky	Covered by
A	Plain $n P_{r}$	just fill $r$ slots	E1
B	Degenerate: $r = 0$ or $r = n$	empty arrangement / all objects; needs $0! = 1$	E2
C	Degenerate: $r > n$	asking the impossible → count is $0$	E2
D	Restricted slot (fill forced first)	a position accepts only some objects	E3
E	Together (glue block)	must remember internal $r!$	E4
F	Apart (gaps method)	count gaps as $k + 1$	E5
G	Fixed ends / both ends forced	two constraints at once	E6
H	"At least" / forbidden (complement)	count total − bad	E7
I	Real-world word problem	translate words → slots	E8
J	Exam twist (mixed patterns + parity)	two restrictions interacting	E9

Every cell A–J gets solved below. If you can do all nine, you have literally seen every shape this topic takes.

E1 — Cell A · Plain $n P_{r}$

Worked example Statement

From the 7 distinct letters ${P, Q, R, S, T, U, V}$ , how many ordered 4-letter strings (no repeats) can be formed?

Forecast: guess whether the answer is bigger or smaller than $7^{4} = 2401$ , and why.

Step 1. We fill 4 positions, left to right, from 7 letters. Why this step? Order matters (a string is ordered), no repeats → this is a permutation, not a combination (Combinations — nCr would ignore order).

Step 2. Count choices per slot: $7, 6, 5, 4$ . Why this step? Each placed letter is used up, so the pool shrinks by one each time — the multiplication principle with dependent choices.

Step 3. Multiply: $7 P_{4} = 7 \cdot 6 \cdot 5 \cdot 4 = 840 = \frac{7 !}{3 !} .$ Why this step? Multiplying is the counting principle; the $\frac{7 !}{3 !}$ form just repackages the same product (the $3! = 3 \cdot 2 \cdot 1$ we "didn't reach" cancels).

Verify: $840 < 2401$ ✓ — fewer than $7^{4}$ because repeats are banned, so each later slot has fewer options. Also $\frac{7 !}{3 !} = \frac{5040}{6} = 840$ ✓.

E2 — Cells B & C · The degenerate inputs

These are the cases students skip and then lose easy marks on. We take all three at once.

Worked example Statement

Evaluate (a) $6 P_{0}$ , (b) $6 P_{6}$ , (c) $4 P_{7}$ .

Forecast: one of these is $1$ , one is $720$ , one is $0$ . Match them before reading.

Step 1 — (a) $6 P_{0}$ . Fill zero slots from 6 objects. Why this step? There is exactly one way to arrange nothing — the empty arrangement. Formally $6 P_{0} = \frac{6 !}{6 !} = 1$ (uses $0! = 1$ from Factorials and 0!). $6 P_{0} = 1.$

Step 2 — (b) $6 P_{6}$ . Arrange all 6 objects. Why this step? Slots get $6, 5, 4, 3, 2, 1$ choices, i.e. $6!$ . In formula form $\frac{6 !}{( 6 - 6 )!} = \frac{6 !}{0 !} = 6!$ . $6 P_{6} = 6! = 720.$

Step 3 — (c) $4 P_{7}$ . Arrange 7 chosen objects out of only 4. Why this step? You'd run out: slots give $4, 3, 2, 1, 0, \dots$ and that $0$ kills the product. There is no valid arrangement. $4 P_{7} = 0.$

Verify: the formula $\frac{4 !}{( 4 - 7 )!} = \frac{4 !}{( - 3 )!}$ is undefined for a factorial of a negative number — matching the common-sense answer "impossible = $0$ arrangements." ✓

E3 — Cell D · Restricted slot, fill the forced position first

Worked example Statement

Arrange all of ${A, B, C, D, E}$ so the arrangement ends in a consonant.

Forecast: more or less than half of all $5! = 120$ arrangements? (Hint: 3 of the 5 letters are consonants.)

Step 1. Handle the last slot first: consonants are ${B, C, D}$ → $3$ choices. Why this step? The constraint lives only on the last position. Locking the tightest slot first guarantees we never leave it unsatisfiable (Pattern 1 / mnemonic "F for Forced first").

Step 2. The remaining 4 letters fill the remaining 4 slots freely: $4! = 24$ . Why this step? Once the last letter is chosen, the other positions have no restriction — plain permutation of what's left.

Step 3. Multiply: $3 \times 24 = 72.$

Verify: by symmetry, fraction ending in a consonant $= \frac{3}{5}$ , so expected count $= \frac{3}{5} \times 120 = 72$ ✓. And $72 > 60$ , i.e. more than half, matching "3 of 5 letters." ✓

E4 — Cell E · Together (glue the block)

Worked example Statement

Six people ${1, 2, 3, 4, 5, 6}$ sit in a row. In how many arrangements are persons $1$ and $2$ next to each other?

Forecast: guess the fraction of all $6! = 720$ seatings in which they end up adjacent.

Step 1. Glue $1$ and $2$ into one block $[12]$ . Now we have 5 items: $[12], 3, 4, 5, 6$ . Why this step? Adjacency means they move as a unit — the gap between them is zero. Treating them as one object bakes "adjacent" into the count automatically (Pattern 2). See the block outlined in teal in the figure.

Step 2. Arrange these 5 items: $5! = 120$ . Why this step? Five distinct items in a row → plain permutation.

Step 3. Inside the block, $1$ and $2$ can swap: $2! = 2$ (orders $12$ and $21$ ). Why this step? "Adjacent" doesn't say who's on the left — both internal orders are valid arrangements. The figure shows the two internal orders side by side.

Step 4. Multiply: $5! \times 2! = 120 \times 2 = 240.$

Verify: fraction adjacent $= \frac{240}{720} = \frac{1}{3}$ — sensible: person 1 has 6 seats; given their seat, person 2's adjacent seats are 2 out of the 5 remaining except at the ends... a quicker sanity check: $\frac{1}{3} \times 720 = 240$ ✓.

E5 — Cell F · Apart (gaps method)

Worked example Statement

Arrange ${A, B, C, D, E}$ in a row so that $A$ , $B$ , and $C$ are no two of them adjacent (all three mutually separated).

Forecast: will slotting three items into gaps leave enough room? Count the gaps first in your head.

Step 1. Arrange the unrestricted letters $D, E$ first: $2! = 2$ . Why this step? The gaps method places the "must-be-apart" objects into the gaps created by the others. So we lay down the free objects first (Pattern 3).

Step 2. Two objects create $2 + 1 = 3$ gaps: $_D_E_$ (shown as plum slots in the figure). Why this step? Gaps $= k + 1$ from $k$ arranged objects — including the two ends. Dropping each of $A, B, C$ into a different gap guarantees none touch.

Step 3. We need to place 3 items into 3 gaps, one per gap: that's $3 P_{3} = 3! = 6$ . Why this step? Exactly 3 gaps and 3 items, all distinct, at most one per gap → ordered placement = a permutation.

Step 4. Multiply: $2! \times 3! = 2 \times 6 = 12.$

Verify: since 3 items into exactly 3 slots forces each slot filled, the pattern is always $D ? E ?$ ... let's list a skeleton: gaps $g_{1} D g_{2} E g_{3}$ , one of $A, B, C$ per gap → $3! = 6$ fillings $\times 2$ orders of $D, E = 12$ ✓. (Note: if we'd required only $A, B$ apart we'd have more room; here the tight count $12$ is correct because gaps exactly match items.)

E6 — Cell G · Both ends fixed (two constraints)

Worked example Statement

Arrange the 5 letters ${A, B, C, D, E}$ so the string starts with a vowel and ends with a consonant.

Forecast: two constraints at once — do they multiply cleanly, or interfere?

Step 1. First slot (vowel): ${A, E}$ → $2$ choices. Why this step? Two forced positions exist; do the constraints one at a time, tightest first. Both ends are equally tight, so start at either.

Step 2. Last slot (consonant): ${B, C, D}$ → but one letter is now used? No — vowels and consonants are disjoint sets, so all 3 consonants are still available → $3$ choices. Why this step? We must check whether Step 1 stole an option from Step 2. Here it didn't (a vowel is never a consonant), so the counts stay independent.

Step 3. Middle 3 slots: 3 letters remain from the 5 → $3! = 6$ . Why this step? After fixing both ends, two letters are used; the leftover 3 fill the middle freely.

Step 4. Multiply: $2 \times 3 \times 3! = 2 \times 3 \times 6 = 36.$

Verify: total arrangements $= 120$ ; fraction $= \frac{36}{120} = \frac{3}{10}$ . Independent-fraction cross-check: P(start vowel) $= \frac{2}{5}$ , P(end consonant | start vowel) $= \frac{3}{4}$ , product $= \frac{6}{20} = \frac{3}{10}$ ✓.

E7 — Cell H · "At least" via complement

Worked example Statement

In how many arrangements of ${A, B, C, D, E}$ do $A$ and $B$ end up together or $A$ is at the start (i.e. at least one of these two conditions holds)?

Forecast: counting overlaps directly is messy — will complement or inclusion–exclusion be cleaner? We use inclusion–exclusion here.

Step 1. Count " $A, B$ together": glue block → $4! \times 2! = 48$ (as in E4's method, but 5 letters). Why this step? This is event $T$ . We already trust the glue method.

Step 2. Count " $A$ at start": fix $A$ first, arrange other 4 → $4! = 24$ . Why this step? This is event $S$ ; one forced position.

Step 3. Count the overlap " $A, B$ together AND $A$ at start": $A$ is at position 1, so $B$ must be at position 2, then $C, D, E$ free → $3! = 6$ . Why this step? When two events can co-occur, $∣ S \cup T ∣ = ∣ S ∣ + ∣ T ∣ - ∣ S \cap T ∣$ — otherwise the overlap is double-counted.

Step 4. Combine: $48 + 24 - 6 = 66.$

Verify: complement check — arrangements with neither condition $= 120 - 66 = 54$ . Direct count of "neither": $A$ not at start (positions 2–5, 4 choices) and $A, B$ not adjacent. Sanity: total not-adjacent $= 120 - 48 = 72$ ; of those, ones with $A$ at start $= 24 - 6 = 18$ ; so not-adjacent AND not-start $= 72 - 18 = 54$ ✓. Both routes give $66$ .

E8 — Cell I · Real-world word problem

Worked example Statement

A photographer lines up 4 medals — gold, silver, bronze, and a special "wildcard" — on a shelf, but the gold and silver medals cannot be next to each other (they'd scratch). How many valid displays?

Forecast: translate to letters. Which method — glue or gaps or complement? Pick before computing.

Step 1. Rename: $G, S, B, W$ ; forbid $G, S$ adjacent. Why this step? Word problems are just permutations wearing costumes; strip the story to distinct objects + a constraint.

Step 2. Total arrangements: $4! = 24$ . Why this step? Four distinct medals, all placed → plain permutation.

Step 3. Count " $G, S$ together" (to subtract): glue → $3! \times 2! = 6 \times 2 = 12$ . Why this step? Complement is easiest here (one forbidden pattern). Glue $G, S$ into a block among $B, W$ = 3 items, times internal $2!$ .

Step 4. Valid $=$ total $-$ together: $24 - 12 = 12.$

Verify: gaps cross-check — arrange $B, W$ first ( $2! = 2$ ), giving $3$ gaps, place $G, S$ in different gaps $= 3 P_{2} = 6$ ; total $2 \times 6 = 12$ ✓. Units make sense: an integer count of displays. ✓

E9 — Cell J · Exam twist (mixed patterns + parity)

Worked example Statement

Using the digits ${1, 2, 3, 4, 5, 6}$ with no repeats, how many 4-digit even numbers greater than 3000 can be formed?

Forecast: two constraints interacting — the leading digit (for ">3000") and the units digit (for "even") both compete for the same small even digits. Guess whether you can multiply naively (you can't).

Step 1 — split on a smart variable. The clash is: an even units digit and a leading digit $\geq 3$ might both want digit $4$ or $6$ . So we case-split on the units digit. Why this step? When two restrictions fight over the same objects, casework on the shared resource keeps the counts clean.

Step 2 — Case (a): units digit $= 2$ . Then thousands digit must be $\in {3, 4, 5, 6}$ (for $> 3000$ , and $\neq = 2$ ) → $4$ choices. Middle two digits from the remaining $4$ digits → $4 \times 3 = 12$ . Why this step? Units is fixed (2 not in the $\geq 3$ set, so no clash with thousands); count thousands, then the two middle slots. $1 \times 4 \times 4 \times 3 = 48.$ Wait — order the slots as (thousands)(mid)(mid)(units): $4$ (thousands) $\times 4 \times 3$ (middles) $= 48$ .

Step 3 — Case (b): units digit $\in {4, 6}$ (2 choices). Now the units digit is one of the $\geq 3$ digits, so it competes with thousands. Thousands must be $\in {3, 4, 5, 6}$ minus the digit already used as units → $4 - 1 = 3$ choices. Middle two from remaining $4$ digits → $4 \times 3 = 12$ . Why this step? Here the earlier choice did remove an option from thousands (unlike E6) — so subtract 1. This is the trap the twist is built around. $2 \times 3 \times 4 \times 3 = 72.$

Step 4 — Add the disjoint cases: $48 + 72 = 120.$ Why this step? The two cases (units $= 2$ vs units $\in {4, 6}$ ) never overlap, so we add (Pattern 5's sibling — partition, then sum).

Verify: sanity — total even 4-digit numbers (no ">3000" rule) $= 3 (units even) \times 5 \times 4 \times 3 = 180$ ; our answer $120 < 180$ as expected since ">3000" removes some. Direct recount of removed ones (thousands $\in {1, 2}$ , even): units even & thousands $\in {1, 2}$ ... gives $60$ , and $180 - 60 = 120$ ✓.

Recall Which cell am I in? (quick triage)

Is $r > n$ ? ::: Answer is $0$ (E2, cell C). Is $r = 0$ or $r = n$ ? ::: Answer is $1$ or $n!$ — remember $0! = 1$ (E2, cells B). One position accepts only some objects? ::: Fill that slot first (E3, cell D). Objects must be adjacent? ::: Glue into a block, times internal $r!$ (E4, cell E). Objects must all be separated? ::: Gaps method, $k + 1$ gaps (E5, cell F). Two forced ends? ::: Fill both, check if they share objects (E6/E9, cells G/J). "At least" / "or" / "not"? ::: Complement or inclusion–exclusion (E7/E8, cells H/I). Two restrictions fight over the same digits? ::: Case-split on the shared resource, then add (E9, cell J).

Connections

Fundamental Counting Principle — every multiply-the-choices step here is this rule.
Combinations — nCr — use when order stops mattering (E1's contrast).
Factorials and 0! — makes the degenerate cells B work ( $0! = 1$ ).
Circular Permutations — the row problems above become $(n - 1)!$ in a ring.
Permutations with Repetition — drop the "no repeats" and counts change.
Probability — Equally Likely Outcomes — the fractions in our "Verify" steps are exactly these probabilities.