Counting with restrictions
Counting with restrictions
Count each admissible outcome once. Decide whether order matters, whether a choice can be repeated, and whether cases overlap before selecting a formula.
| Restriction | Strategy | Example |
|---|---|---|
| Repeated choices permitted | Multiply choices for each position. | Four-digit codes from \(0\)–\(9\), allowing leading zero: \(10^4\). |
| Two distinct people together | Treat the pair as a block, then arrange inside it. | Six distinct people in a row: \(5!\times2!\). |
| At least one chosen | Subtract the case with none. | Three people from four tutors and five students, with a tutor: \(\binom93-\binom53=74\). |
Example
Choose two students from five and one tutor from four: \(\binom52\binom41=40\). Giving those three people distinct roles would add an arrangement step; a committee has no such step.
Use disjoint cases for “exactly”, “at most” or “at least” restrictions. When using a complement, count the total and excluded outcomes in the same sample space.
For \(r\) objects with \(x_1,\ldots,x_k\) identical objects of each type, where \(x_1+\cdots+x_k=r\), the number of distinct arrangements is \(\dfrac{r!}{x_1!\cdots x_k!}\).
If \(k\) specified objects among \(r\) distinct objects must be consecutive, treat them as one block: \(k!(r-k+1)!\). If no two of those \(k\) objects may be adjacent, arrange the other \(r-k\) objects, choose \(k\) of the \(r-k+1\) gaps and arrange the specified objects: \((r-k)!\binom{r-k+1}{k}k!\), provided \(k\le r-k+1\); otherwise there are no arrangements.
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