Counting & ProbabilityDifficulty 10-750 tagged problems

Constructive Counting

Constructive Counting is the habit of counting objects by building them step-by-step with controlled choices. 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.

No tagged problems yet We are still curating launch-ready practice for this technique.

The Key Result

$$\#(\text{outcomes}) = c_1 \cdot c_2 \cdots c_m, \quad c_i = \#\text{choices at step } i$$

Build each outcome one decision at a time and multiply the number of independent choices at every step.

Why It Works

1
If a process has $m$ stages and stage $i$ offers $c_i$ choices no matter what was chosen before, the choices combine independently.
2
Each complete outcome is a unique sequence of choices, so the total is the product $c_1 c_2 \cdots c_m$ (the multiplication principle).
3
The key requirement: the count $c_i$ at each step must not depend on the specific earlier picks (only on which step you are at).
4
Pick the order of construction that keeps the per-step counts constant — that is what makes the product valid.

Worked Examples

Example 1

A license plate has $3$ letters (A–Z) followed by $3$ digits (0–9), repeats allowed. How many plates exist?

1
Build the plate slot by slot. Each of the $3$ letter slots has $26$ choices.
2
Each of the $3$ digit slots has $10$ choices, independent of the letters.
3
Multiply: $26 \cdot 26 \cdot 26 \cdot 10 \cdot 10 \cdot 10 = 26^3 \cdot 10^3$.
4
Compute: $17576 \cdot 1000 = 17576000$.

Answer:

$17576000$

Warm-up

How many $3$-digit numbers (leading digit $1$–$9$) have all distinct digits?

1
Build digit by digit. The hundreds digit is nonzero: $9$ choices.
2
The tens digit must differ from the hundreds digit but may be $0$: of the $10$ digits, $9$ remain.
3
The units digit differs from both prior digits: $8$ remain.
4
Multiply: $9 \cdot 9 \cdot 8 = 648$.

Answer:

$648$

Contest level

How many ways can $4$ people be seated in a row of $4$ chairs if two specific people, Alice and Bob, must sit next to each other?

1
Glue Alice and Bob into a single block; now arrange $3$ items (the block plus the other $2$ people) in a row: $3! = 6$ ways.
2
Inside the block, Alice and Bob can be in either order: $2$ ways.
3
These two choices are independent, so multiply: $6 \cdot 2 = 12$.

Answer:

$12$

Olympiad / Challenge

How many ways can $8$ rooks be placed on an $8 \times 8$ chessboard so that no two attack each other (no two share a row or column)?

1
No two rooks share a row, and there are $8$ rooks and $8$ rows, so exactly one rook sits in each row.
2
Build the placement row by row. The rook in row $1$ may go in any of $8$ columns.
3
The rook in row $2$ must avoid that column: $7$ columns left. Row $3$: $6$ columns. And so on down to row $8$: $1$ column.
4
Multiply: $8 \cdot 7 \cdot 6 \cdots 1 = 8! = 40320$.

Answer:

$40320$

Going Deeper

The multiplication principle generalizes to any process where the number of valid choices at each stage is constant regardless of the earlier (specific) picks — even if the available items change, as long as the count does not. The non-attacking-rooks count being $8!$ is the same reason a permutation has $n!$ orderings: each step removes exactly one option.
Constructive counting is the default first approach for nearly all 'how many ways' problems: forming numbers digit by digit, seating people, building strings, assigning tasks. Pairing it with the 'glue together' trick (treat must-be-adjacent items as one block) handles a huge class of arrangement constraints.
Pitfall: the per-step choice count must not depend on which specific earlier choices were made — only on the stage. If choosing item A early changes how many options a later step has (in a way that varies case by case), the simple product fails and you must split into cases first. Build in the order that keeps each step's count uniform, and put the most constrained slot first.

Spot the Signal

Look for problems where the key step is counting objects by building them step-by-step with controlled choices.
You can describe the hard part as counting objects by building them step-by-step with controlled choices, 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 constructive counting, then rewrite the givens around it.
Name the relation that makes Constructive Counting 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 Constructive Counting just because the surface looks familiar; verify the required condition first.
Applying Constructive Counting 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 constructive counting reveal the counting structure; then compute only what remains.

Try it on:

Practice a contest problem where the key step is counting objects by building them step-by-step with controlled choices.