Permutations & Combinations

Q: What is $0!$?

By definition $0! = 1$. That keeps the formulas working at the edges: $\dbinom{n}{n} = \dfrac{n!}{n!\,0!} = 1$, which is right — there's exactly one way to choose everything.

Q: Which is bigger, $_nP_r$ or $_nC_r$?

The permutation, whenever $r > 1$. Since $_nP_r = r! \cdot {_nC_r}$, there are always at least as many arrangements as selections — each selection can be arranged $r!$ ways.

Counting problems come in two flavors. A permutation counts arrangements, where order matters: gold-silver-bronze is different from bronze-silver-gold. A combination counts selections, where order does not matter: a committee of Ana, Ben, and Cruz is the same committee no matter who was picked first.

One question decides which tool you need: if you swapped two of the chosen items, would you get a genuinely different result? Yes means permutation. No means combination. Everything after that is plugging into a formula.

The two formulas

Both formulas use factorials, where $n! = n(n-1)(n-2)\cdots 1$ . Arranging $r$ items chosen from $n$ : $_nP_r = \dfrac{n!}{(n-r)!}$ . Selecting $r$ items from $n$ when order doesn't matter: $_nC_r = \dfrac{n!}{r!(n-r)!}$ , which is also written $\dbinom{n}{r}$ .

Notice the only difference is the extra $r!$ in the combination's denominator. That $r!$ divides out the repeats: every group of $r$ items shows up $r!$ times among the ordered arrangements, once for each way to shuffle it.

How to compute them quickly

You rarely need the full factorials. For $_7P_3$ , just multiply $3$ falling numbers starting at $7$ : $7 \cdot 6 \cdot 5 = 210$ . For $\dbinom{7}{3}$ , take that same product and divide by $3! = 6$ : $\dfrac{210}{6} = 35$ .

For combinations, cancel before you multiply. In $\dfrac{7 \cdot 6 \cdot 5}{3 \cdot 2 \cdot 1}$ , the $6$ on top cancels the $3 \cdot 2$ on the bottom, leaving $7 \cdot 5 = 35$ with no big numbers ever appearing.

Spotting which one a problem wants

Permutation words: arrange, line up, order, schedule, ranking, president-then-vice-president (distinct roles). Combination words: choose, select, committee, group, handful, toppings — anything where the chosen items end up in one unlabeled pile.

The most common error is using ordered counting for an unordered selection. Choosing $3$ committee members from $10$ by computing $10 \cdot 9 \cdot 8 = 720$ counts every committee $3! = 6$ times. The correct count is $\dfrac{720}{6} = 120$ .

Worked examples

Example 1: a permutation (order matters)

In how many ways can $4$ people line up in a row?

A line-up is an arrangement, so order matters

All

4

people are used, so this is

4!

_4P_4 = 4!

Multiply

4 \cdot 3 \cdot 2 \cdot 1 = 24

Answer: $24$ ways

Example 2: a combination (order doesn't matter)

How many ways can $3$ items be chosen from $6$ when order does not matter?

A selection with no order is a combination

_6C_3 = \dfrac{6!}{3!\,3!}

Write it as a product over

3!

\dfrac{6 \cdot 5 \cdot 4}{3 \cdot 2 \cdot 1}

Cancel and multiply

\dfrac{120}{6} = 20

Answer: $20$ ways

Example 3: a partial permutation

How many ways can $2$ letters be arranged from $4$ different letters when order matters?

Order matters, so use a permutation

_4P_2 = \dfrac{4!}{2!}

Multiply

2

falling numbers starting at

4

4 \cdot 3 = 12

Answer: $12$ ways

Try one yourself

Common questions

How do I decide between a permutation and a combination?

Ask whether swapping two chosen items creates a different result. Different seats, ranks, or roles mean order matters — permutation. One unlabeled group means order does not matter — combination.

Why does the combination formula divide by

r!

Ordered counting lists every group of $r$ items $r!$ times — once per shuffle of that group. Dividing by $r!$ collapses those repeats so each group is counted exactly once.

What is

0!

By definition $0! = 1$ . That keeps the formulas working at the edges: $\dbinom{n}{n} = \dfrac{n!}{n!\,0!} = 1$ , which is right — there's exactly one way to choose everything.

Which is bigger,

_nP_r

_nC_r

The permutation, whenever $r > 1$ . Since $_nP_r = r! \cdot {_nC_r}$ , there are always at least as many arrangements as selections — each selection can be arranged $r!$ ways.

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