How Many Ways Can Five Friends Sit in a Row? Counting Without Listing

Listing every arrangement quickly becomes impossible: ten books on a shelf can be arranged in more than three million ways. Counting techniques find such totals in a few lines.

The key ideas are to break a task into stages and multiply, and to use factorials as a compact way of writing long products.

This part covers the fundamental principle of counting, factorials, permutations of distinct objects, and permutations when some objects are alike.

**If one stage can be done in $m$ ways and, after it, the next can be done in $n$ ways, the two stages together can be done in $m\times n$ ways; the rule extends to any number of stages.

Worked example 1.** With $3$ shirts and $4$ trousers, the number of outfits is $3\times 4=12$.

Worked example 2. A $4$-digit PIN uses digits $0$ to $9$ with repetition allowed:

$$10\times 10\times 10\times 10=10000$$

Worked example 3. Three-digit numbers with distinct digits chosen from $1$ to $9$:

$$9\times 8\times 7=504$$

Worked example 4. Signals are made by hanging $2$ of $5$ different flags one below the other:

$$5\times 4=20\text{ signals}$$

An everyday example. **A dosa counter offering $3$ types of dosa and $4$ chutneys**, one of each, gives $3\times 4=12$ possible plates.

The substance. Multiply when choices happen together; add when they are alternatives — going by one of $3$ buses or one of $2$ trains gives $3+2=5$ ways.

**$n!=n\times (n-1)\times \cdots \times 2\times 1$ for a natural number $n$, with $0!=1$; ratios of factorials simplify by cancelling the common tail.**

$$5!=1206!=72010!=3628800$$

Worked example 1.

$$\frac{7!}{5!}=\frac{7\times 6\times 5!}{5!}=42\frac{8!}{6!2!}=\frac{8\times 7}{2}=28$$

Worked example 2. Find $n$ if $\frac{n!}{(n-2)!}=12$.

$$n(n-1)=12n=4$$

Worked example 3. Find $x$ if $\frac{1}{6!}+\frac{1}{7!}=\frac{x}{8!}$. Multiply by $8!$:

$$\frac{8!}{6!}+\frac{8!}{7!}=56+8=64x=64$$

An everyday example. **Arranging $10$ different books on a shelf** can be done in $10!=3628800$ ways.

The substance. **$0!=1$ is defined so that the formulas still work** when every object is used, such as $\frac{n!}{(n-n)!}=n!$.

**Arranging $r$ of $n$ distinct objects without repetition fills the first place in $n$ ways, the next in $n-1$, and so on, giving ${}^n{P}_r=n(n-1)\dots (n-r+1)=\frac{n!}{(n-r)!}$; with repetition allowed, the count is $n^r$.

Worked example 1.** Five friends in a row: ${}^5{P}_5=5!=120$. Any $3$ of them in a row: ${}^5{P}_3=\frac{5!}{2!}=60$.

Worked example 2. Four-digit numbers from digits $1$ to $9$ with no repeats:

$${}^9{P}_4=9\times 8\times 7\times 6=3024$$

Worked example 3 — with repetition. Three-letter strings from $5$ letters, repeats allowed: $5^3=125$.

Worked example 4 — a condition. Three-digit even numbers from $1,2,3,4,6,7$, no repetition. The units digit must be $2$, $4$ or $6$:

$$3\times (5\times 4)=60$$

An everyday example. **At a school sports day with $8$ runners**, gold, silver and bronze can go to them in ${}^8{P}_3=8\times 7\times 6=336$ ways.

The substance. Fill the restricted place first — here the units digit — then count the rest.

**If $n$ objects include $p_1$ alike of one kind, $p_2$ alike of another, and so on, the number of distinct arrangements is $\frac{n!}{p_1!p_2!\dots }$.

Why.** Swapping identical objects produces no new arrangement, so each distinct arrangement has been counted $p_1!p_2!\dots$ times among the $n!$.

Worked example 1. INDIA has $5$ letters with $I$ twice:

$$\frac{5!}{2!}=60$$

Worked example 2. ALLAHABAD has $9$ letters: $A$ four times, $L$ twice.

$$\frac{9!}{4!2!}=\frac{362880}{48}=7560$$

Worked example 3. MISSISSIPPI has $11$ letters: $I$ four, $S$ four, $P$ two.

$$\frac{11!}{4!4!2!}=34650$$

**With all four $I$s together**, treat them as one block: $8$ objects with $S$ four and $P$ two, giving $\frac{8!}{4!2!}=840$.

An everyday example. **Placing $3$ identical red diyas and $2$ identical white diyas in a row** gives $\frac{5!}{3!2!}=10$ different patterns.

The substance. Only identical objects are divided out; distinct letters such as $H$, $B$ and $D$ contribute $1!$.