Permutations vs combinations: telling them apart

Students lose more marks in this chapter to picking the wrong tool than to any arithmetic. And the tool is chosen by asking one plain question of the situation: does the order matter?

If rearranging the same items gives you a genuinely different outcome, order matters, and you are counting permutations. If rearranging them gives you the same outcome, order does not matter, and you are counting combinations.

A quick example makes the difference vivid. Suppose you pick 3 students from a class.

If you are choosing a president, a secretary, and a treasurer, then Aisyah-as-president / Ben-as-secretary is a different result from Ben-as-president / Aisyah-as-secretary, the roles give the order meaning, so this is a permutation. But if you are just choosing 3 students to form a study group with no roles, then {Aisyah, Ben, Chong} is the same group however you list them, order is irrelevant, so this is a combination.

Both formulas are built from the factorial. The factorial of a whole number $n$, written $n!$, is the product of every whole number from $n$ down to $1$:

$$n!=n\times(n-1)\times(n-2)\times\cdots\times 2\times 1$$

So $5!=5\times 4\times 3\times 2\times 1=120$, and $3!=6$. One value surprises many students: $0!=1$, by definition, it makes the formulas behave correctly, so simply accept it.

A factorial counts the number of ways to arrange a set of distinct objects in a row: $n$ different objects can be lined up in $n!$ ways, because there are $n$ choices for the first place, then $n-1$ for the second, and so on.

That idea, multiplying the choices at each stage, is the multiplication principle, and it quietly underpins the whole chapter.

A permutation counts arrangements, where order counts. The number of ways to arrange $r$ items chosen from $n$ distinct items is:

$${}^{n}P_{r}=\frac{n!}{(n-r)!}$$

A combination counts selections, where order does not count. The number of ways to choose $r$ items from $n$ distinct items is:

$${}^{n}C_{r}=\frac{n!}{r!\,(n-r)!}$$

Look closely and you will see the combination formula is just the permutation formula divided by an extra $r!$. That extra $r!$

is the number of ways to reorder the chosen $r$ items, and dividing by it is exactly how we throw away the orderings we no longer care about. This is why ${}^{n}C_{r}$ is always smaller than (or equal to) ${}^{n}P_{r}$: every unordered selection corresponds to several ordered arrangements.

Using the same figures for both makes the contrast sharp. Suppose there are 8 members in a club.

First, the club must choose a chairperson, a secretary, and a treasurer, three distinct roles. Order matters, so this is a permutation: ${}^{8}P_{3}=\frac{8!}{5!}=8\times 7\times 6=336$ ways.

The shortcut is worth internalising: ${}^{8}P_{3}$ is simply the product of the top 3 numbers counting down from 8.

Now, instead, the club must choose 3 members for an unnamed committee, no roles, everyone equal. Order does not matter, so this is a combination: ${}^{8}C_{3}=\frac{8!}{3!\,5!}=\frac{8\times 7\times 6}{3\times 2\times 1}=\frac{336}{6}=56$