Math Core

Many counting problems have no formula that fits. The reliable fallback is to split the problem into cases and count each case with an organized list. Casework shows up on almost every MATHCOUNTS and AMC 8 test, and it is the tool you reach for when you aren't sure what else to do.

Organized lists

An organized list writes the possibilities in a fixed order so that you never skip one and never write one twice. The trick is to decide in advance which quantity you change first.

Suppose you want every way to make $25$ cents from pennies, nickels and dimes. Changing coins at random is a recipe for mistakes. Instead, fix the number of dimes (the biggest coin) first, then the number of nickels, and let pennies fill in the rest.

Dimes	Nickels	Pennies
$2$	$1, 0$	$0, 5$
$1$	$3, 2, 1, 0$	$0, 5, 10, 15$
$0$	$5, 4, 3, 2, 1, 0$	$0, 5, \dots, 25$

That gives $2 + 4 + 6 = 12$ ways. Once the dimes and nickels are chosen, the pennies are forced, so each row of the list really is one way.

Casework

Casework is the same idea at a larger scale: break the whole count into smaller counts you can handle, then add.

Rules for good casework

Choose the cases by one feature (the first digit, the largest part, the number of a certain coin).
The cases must not overlap: no possibility may be counted in two cases.
The cases must cover everything: every possibility belongs to some case.
Count each case separately, then add.

A good case split makes each case easy. Splitting on the most restrictive feature (the biggest coin, the largest side, the leading digit) usually works best, because it leaves the fewest choices for everything else.

Worked example: Digit sums

How many three-digit numbers have digits that add up to $5$ ?

Split on the hundreds digit $a$ , which can be $1$ to $5$ (it can't be $0$ ). The other two digits must add to $5 - a$ . Two digits that add to $s$ can be chosen in $s + 1$ ways ( $0 + s$ , $1 + (s-1)$ , …, $s + 0$ ).

$a$	tens + units	ways
$1$	$4$	$5$
$2$	$3$	$4$
$3$	$2$	$3$
$4$	$1$	$2$
$5$	$0$	$1$

Total: $5 + 4 + 3 + 2 + 1 = 15$ .

Worked example: Triangles with a fixed perimeter

How many non-congruent triangles have integer side lengths and perimeter $12$ ?

List the sides in order $a \le b \le c$ so each triangle is written only once. The triangle inequality says $a + b > c$ . Since $a + b = 12 - c$ , that means $12 - c > c$ , so $c < 6$ . Also $c$ is the largest side, so $c \ge 4$ .

$c = 5$ : $a + b = 7$ with $a \le b \le 5$ : $(2, 5)$ and $(3, 4)$ .
$c = 4$ : $a + b = 8$ with $a \le b \le 4$ : $(4, 4)$ .

The triangles are $2\text{-}5\text{-}5$ , $3\text{-}4\text{-}5$ and $4\text{-}4\text{-}4$ : $3$ triangles.

Worked example: Counting by the last step

In how many ways can you climb a staircase of $6$ steps if each move goes up $1$ or $2$ steps?

Let $f(n)$ be the number of ways to climb $n$ steps. Split on the last move. If it is a $1$ -step, the moves before it climb $n - 1$ steps; if it is a $2$ -step, they climb $n - 2$ steps. The cases don't overlap and cover everything, so

$f(n) = f(n-1) + f(n-2).$

With $f(1) = 1$ and $f(2) = 2$ : $f(3) = 3$ , $f(4) = 5$ , $f(5) = 8$ , $f(6) = 13$ . There are $13$ ways.

Common mistake

The most common casework error is overlapping cases. If you count "numbers with a $3$ " and "numbers with a $5$ " as separate cases and add, you count $35$ twice. Pick a single feature to split on so each possibility lands in exactly one case.

Tip

Before adding, check one or two cases by writing them out fully. If a case seems to have a pattern (like $5, 4, 3, 2, 1$ above), confirm the first and last entries and trust the pattern in between.

Practice

Practice 1

In how many ways can you make $30$ cents using nickels, dimes and quarters? (You don't have to use every kind of coin.)