Exercises — Permutations — nPr, arrangements with restrictions

WHAT we do: fill $3$ slots from $7$ distinct objects. WHY these numbers: slot 1 has all $7$ available; once one is used, slot 2 has $6$; then slot 3 has $5$. $7P_3=7\times 6\times 5=210$ Check against the compact form: $\frac{7!}{(7-3)!}=\frac{7!}{4!}=\frac{5040}{24}=210.$ ✓

Arranging everything means $r=n$, so we want $6P_6=\frac{6!}{0!}$. WHY $0!=1$: there is exactly one way to arrange nothing, so dividing by $0!$ changes nothing (see Factorials and 0!). $6P_6=6!=720$

The test: does swapping the two chosen people give a different outcome? Yes — (Ana President, Bo Secretary) $\ne$ (Bo President, Ana Secretary). Order matters $\Rightarrow$ permutation, not Combinations — nCr. $10P_2=10\times 9=90$

WHAT restricts: "odd" only touches the units digit, so lock that slot first (tightest constraint first).

• Units: must be odd $\Rightarrow \{1,3,5\}\Rightarrow 3$ choices.
• Thousands: any of the remaining $5$ $\Rightarrow 5$.
• Hundreds: $4$ remaining.
• Tens: $3$ remaining. $3\times 5\times 4\times 3=180$

Glue the $3$ friends into one block $[FFF]$. Now we arrange $4$ items: the block plus the other $3$ people.

• Arrange the $4$ items: $4!=24$.
• Inside the block the $3$ friends reorder among themselves: $3!=6$. $4!\times 3!=24\times 6=144$

Gaps method. First arrange the other $3$ letters $R,S,T$: that's $3!=6$. Look at the figure: three placed letters create $\underline{}R\underline{}S\underline{}T\underline{}$ — that is $3+1=4$ gaps.

WHY complement: "at least one even" spans many cases (one even, two even, three even). Its opposite — "no even digit at all" — is a single clean case.

• Total $3$-digit numbers: $7P_3=7\times 6\times 5=210$.
• All-odd numbers use only $\{1,3,5,7\}$ (the $4$ odds): $4P_3=4\times 3\times 2=24$. $\text{at least one even}=210-24=186$

Lock the two forced slots first, then fill the middle.

• First slot (vowel): $3$ choices.
• Last slot (consonant): $3$ choices.
• Middle $4$ slots: the remaining $4$ letters in any order $\Rightarrow 4!=24$. $3\times 3\times 4!=9\times 24=216$

The hidden restriction: the leading digit can't be $0$, so lock the thousands slot first.

• Thousands: any of $\{1,2,3,4\}$ $\Rightarrow 4$ choices (0 forbidden here).
• Now $0$ is back in play for the rest. Remaining slots draw from the $4$ leftover digits:
• Hundreds: $4$ choices.
• Tens: $3$.
• Units: $2$. $4\times 4\times 3\times 2=96$

Glue then subtract. Treat $[AB]$ as a block; the block gives $4!\times 2!=48$ total adjacent arrangements (parent E3). Now remove those where $A$ sits at an end. Look at the two orderings of the block:

• Block $=AB$: $A$ is at an end only if the block occupies the leftmost position (so $A$ is position 1). That's $1$ block-placement $\times 3!$ for the others $=6$.
• Block $=BA$: $A$ is at an end only if the block occupies the rightmost position (so $A$ is position 5). Again $3!=6$. Forbidden (A at an end, adjacent) $=6+6=12$. $48-12=36$

Gaps method, done in order.

• Arrange the $4$ boys first: $4!=24$.
• Four boys create $4+1=5$ gaps: $\underline{}B\underline{}B\underline{}B\underline{}B\underline{}$.

  

• Place $3$ girls into $3$ different gaps (different gaps $\Rightarrow$ never adjacent), order matters: $5P_3=5\times 4\times 3=60$. $4!\times 5P_3=24\times 60=1440$

Complement is cleaner: "not all together" $=$ total $-$ "all three together".

• Total: $6!=720$.
• All three together (glue $[ABC]$): $4!\times 3!=24\times 6=144$. $720-144=576$

Two restrictions collide: units must be even $\{0,2,4\}$, and thousands $\ne 0$. When the units grabs $0$, it changes what the thousands slot may use — so split on whether the units digit is $0$.

Case A — units $=0$:

• Units: $1$ way ($0$).
• Thousands: any of remaining $\{1,2,3,4,5\}$ $\Rightarrow 5$ (no zero conflict, $0$ already used).
• Hundreds: $4$ remaining.
• Tens: $3$ remaining.
• Subtotal $=1\times 5\times 4\times 3=60$.

Case B — units $\in \{2,4\}$:

• Units: $2$ ways.
• Thousands: can't be $0$ and can't equal the units digit. From $6$ digits remove $0$ and the used even $\Rightarrow 4$ choices.
• Hundreds: $4$ remaining (now $0$ is allowed here).
• Tens: $3$ remaining.
• Subtotal $=2\times 4\times 4\times 3=96$.

$60+96=156$

Two glue blocks.

• Glue the $3$ Maths into block $M$, the $2$ Physics into block $P$. Items to arrange: $M,P,C_1,C_2$ $\Rightarrow 4$ items $\Rightarrow 4!=24$.
• Inside $M$: $3!=6$. Inside $P$: $2!=2$. $4!\times 3!\times 2!=24\times 6\times 2=288$

This links straight to Probability — Equally Likely Outcomes: all $6!$ orderings are equally likely, so probability $=\frac{\text{favourable}}{\text{total}}$.

• Total: $6!=720$.
• Favourable: both ends are vowels. There are exactly $2$ vowels $\{A,E\}$ and $2$ end-slots. Place them: $2P_2=2!=2$. Middle $4$ letters: $4!=24$. Favourable $=2\times 24=48$. $P=\frac{48}{720}=\frac{1}{15}$

Every one of these is the same move: line up empty slots, and at each slot ask "how many are still allowed right now?" then multiply. Restrictions just change one of those counts. When two restrictions push on the same slot (even and no leading zero), split into cases so each case has clean counts.

L1: $210$, $720$, $90$. L2: $180$, $144$, $72$. L3: $186$, $216$, $96$. L4: $36$, $1440$, $576$. L5: $156$, $288$, $\frac{1}{15}$.