Why Seating People Around a Table Gives Fewer Arrangements Than a Row

How many PIN codes, number plates or seating plans are possible? Listing them is hopeless once the numbers grow, but a few counting principles give the answer in a line. Permutations count arrangements in which order matters.

This lesson covers the fundamental principle of counting and factorials, the nPr formula with restrictions, arranging digits and letters, and circular permutations.

**The fundamental principle of counting states that if one task can be done in m ways and a second in n ways, the two can be done one after the other in $m\times n$ ways, and $n!=n(n-1)(n-2)\cdots 2\times 1$ counts the ways to arrange n different objects in a row.

Multiplication and addition:

- And** — one task after another — multiply: $m\times n$
- Or — one task or the other — add: $m+n$

Factorials. $0!=1$, $5!=120$, and $n!=n\times (n-1)!$.

Worked example. A school campus has 4 paths from the gate to the library and 3 from the library to the laboratory. A student can walk from the gate to the laboratory through the library in $4\times 3=12$ ways.

An everyday example. A vehicle registration series with 2 letters followed by 4 digits allows $26^2\times 10^4=6760000$ different number plates.

The substance. **$0!=1$ is not an arbitrary rule** — it keeps the formula $\frac{n!}{(n-r)!}$ correct when $r=n$.

**The number of arrangements of r objects chosen from n different objects is ${}^n{P}_r=\frac{n!}{(n-r)!}$, because the first place can be filled in n ways, the second in $n-1$ ways, and so on down to $n-r+1$; items that must stay together are tied into one block, and 'never together' is the total minus 'always together'.

Derivation.**

$${}^n{P}_r=n(n-1)(n-2)\cdots (n-r+1)=\frac{n!}{(n-r)!}$$

Worked example. The posts of head, deputy and sports captain are filled from 10 students in ${}^{10}{P}_3=10\times 9\times 8=720$ ways.

Always together. Arrange 6 books on a shelf so that 2 particular books are always together. Treat the pair as one block: $5!$ arrangements of the blocks and $2!$ orders inside the pair give $120\times 2=240$.

Never together. There are $6!=720$ arrangements in all, so the 2 books are never together in $720-240=480$ ways.

An everyday example. Filling the first 3 places in a cricket batting order from 11 players gives ${}^{11}{P}_3=990$ possibilities.

The substance. Always multiply by the arrangements inside a block — forgetting the $2!$ is the commonest error.

**Without repetition the number of choices falls by one at each position, with repetition every position has the full set of choices, and when some letters are identical the count is $\frac{n!}{p!q!r!}$, dividing out the rearrangements of each repeated group.

Worked example (digits). How many 3-digit numbers can be formed from 0, 1, 2, 3, 4?

- Without repetition** — the hundreds place cannot be 0: $4\times 4\times 3=48$
- With repetition — $4\times 5\times 5=100$
- Even numbers without repetition — ending in 0: $4\times 3=12$; ending in 2 or 4: $2\times 3\times 3=18$; total $30$

Worked example (letters). MISSISSIPPI has 11 letters, with I four times, S four times and P twice:

$$\frac{11!}{4!4!2!}=\frac{39916800}{24\times 24\times 2}=34650$$

An everyday example. The letters of the word INDIA can be arranged in $\frac{5!}{2!}=60$ ways, since I appears twice.

The substance. Fill the restricted position first — choosing the hundreds digit before the others stops numbers beginning with 0 from being counted.

**Arranging n different objects in a circle gives $(n-1)!$ ways, because rotations of one arrangement are identical; when clockwise and anticlockwise orders cannot be told apart, as with beads on a necklace, the count halves to $\frac{(n-1)!}{2}$.

Why $(n-1)!$.** Fix one object's position to remove rotations, then arrange the remaining $n-1$ objects in $(n-1)!$ ways.

Distinguishable or not:

- People around a table — clockwise and anticlockwise differ: $(n-1)!$
- Beads on a necklace or flowers in a garland — turning it over gives the same arrangement: $\frac{(n-1)!}{2}$

Worked example. 6 friends sit around a round table in $(6-1)!=120$ ways. If 2 of them insist on sitting together, treat them as one unit: $(5-1)!\times 2!=48$ ways.

Worked example 2. 7 different beads make a bracelet in $\frac{6!}{2}=360$ ways.

An everyday example. Guests seated at a round table at a family function follow the $(n-1)!$ rule, while a marigold garland of different flowers follows the halved rule.

The substance. Numbered seats remove the symmetry — if the chairs are labelled, the arrangement behaves like a row and the count returns to $n!$.