Casework is the habit of splitting a problem into complete non-overlapping cases that are easier to solve. In contest math, that habit turns a crowded setup into a relation the student can test, bound, count, or compute. MathGrit teaches it as a recognizable signal, a deliberate move, and a final translation back to the original question.
Split a count into separate cases that never overlap and together cover everything, count each case, then add the results.
Why It Works
1
If a set $S$ is partitioned into disjoint subsets $S_1, S_2, \ldots, S_k$ (no element in two cases), then $|S| = |S_1| + |S_2| + \cdots + |S_k|$.
2
The two requirements are exhaustive (every outcome lands in some case) and mutually exclusive (no outcome lands in two cases) — otherwise you undercount or double-count.
3
Choose the splitting variable that makes each case simple: a digit's value, how many objects are of one type, or which item is largest.
4
Count each case independently with whatever tool fits, then sum.
Worked Examples
Example 1
How many ordered pairs $(a, b)$ of dice rolls (each from $1$ to $6$) have $a + b = 7$?
1
Split into cases by the value of $a$, since $b = 7 - a$ is then forced.
2
$a = 1 \Rightarrow b = 6$; $a = 2 \Rightarrow b = 5$; $a = 3 \Rightarrow b = 4$; $a = 4 \Rightarrow b = 3$; $a = 5 \Rightarrow b = 2$; $a = 6 \Rightarrow b = 1$.
3
Each of these $6$ cases gives exactly one valid pair, and no pair is counted twice.
4
Add them: $1 + 1 + 1 + 1 + 1 + 1 = 6$.
Answer:
$6$
Warm-up
How many $4$-digit integers (so the leading digit is $1$–$9$) have digits that are all even?
1
Split the construction by position. The thousands digit must be even AND nonzero, so it is one of $\{2,4,6,8\}$: $4$ choices.
2
Each of the remaining three digits (hundreds, tens, units) may be any even digit $\{0,2,4,6,8\}$: $5$ choices each.
3
These cases are independent across positions, so multiply: $4 \cdot 5 \cdot 5 \cdot 5$.
4
Compute: $4 \cdot 125 = 500$.
Answer:
$500$
Contest level
How many ways can you make $25$ cents using pennies ($1$), nickels ($5$), and dimes ($10$), where order does not matter?
1
Case on the number of dimes $d$, since that is the most restrictive coin. We need $10d \le 25$, so $d \in \{0, 1, 2\}$.
2
$d = 0$: must make $25$ from nickels and pennies. The number of nickels $n$ satisfies $5n \le 25$, so $n \in \{0,1,2,3,4,5\}$ — $6$ ways (pennies fill the rest).
3
$d = 1$: make $15$ from nickels and pennies, $5n \le 15$ gives $n \in \{0,1,2,3\}$ — $4$ ways.
4
$d = 2$: make $5$ from nickels and pennies, $5n \le 5$ gives $n \in \{0,1\}$ — $2$ ways.
5
Cases are disjoint (fixed $d$) and exhaustive; add them: $6 + 4 + 2 = 12$.
Answer:
$12$
Olympiad / Challenge
How many $5$-letter strings over $\{A, B, C\}$ contain no two adjacent equal letters and use the letter $A$ at least once?
1
First count all strings with no two adjacent equal letters: the first letter has $3$ choices, and each subsequent letter must differ from its predecessor, giving $2$ choices. So $3 \cdot 2^4 = 48$.
2
Use complementary counting within this set: subtract those using no $A$, i.e. strings over $\{B, C\}$ with no two adjacent equal.
3
Over a $2$-letter alphabet with no two adjacent equal, the string alternates, so the first letter ($2$ choices) forces all others: $2$ such strings of length $5$ ($BCBCB$ and $CBCBC$).
4
Subtract: $48 - 2 = 46$.
Answer:
$46$
Going Deeper
Casework is the universal fallback: any count can be split by some controlling variable (largest element, number of objects of a type, a digit's value). The art is choosing the split that makes each case independent and uniform so you can multiply or reuse a formula, rather than splitting into so many cases that the bookkeeping itself errs.
It appears whenever a constraint behaves differently across ranges — a leading digit barred from $0$, a coin that caps how many of another fit, a value forcing the rest. On AMC/AIME it is often the cleanest route and is frequently combined with complementary counting inside a single case.
Pitfall: cases must be both mutually exclusive and exhaustive. The two classic failures are double-counting (an outcome lands in two cases, e.g. a string counted under both 'starts with A' and 'ends with A') and a missed case (forgetting $d = 0$, or the empty configuration). Always state your splitting variable explicitly and check the case ranges cover every value with no overlap.
Spot the Signal
Look for problems where the key step is splitting a problem into complete non-overlapping cases that are easier to solve.
You can describe the hard part as splitting a problem into complete non-overlapping cases that are easier to solve, but a direct attack starts producing clutter.
The problem rewards preserving structure instead of expanding, listing, or guessing too early.
Learn the Move
Start by identify the counting object that calls for casework, then rewrite the givens around it.
Name the relation that makes Casework legal before doing computation.
Use the new relation to replace the messiest part of the problem with a cleaner one.
Translate the result back to the quantity the problem actually asks for.
Avoid These Traps
Do not use Casework just because the surface looks familiar; verify the required condition first.
Applying Casework because it sounds relevant, without checking the trigger first.
Stopping after spotting the technique instead of finishing the calculation or proof.
MathGrit Coach Note
Let casework reveal the counting structure; then compute only what remains.
Try it on:
Practice a contest problem where the key step is splitting a problem into complete non-overlapping cases that are easier to solve.