Permutations vs Combinations

Permutations $P(n, r) = \frac{n!}{(n-r)!}$ count ordered arrangements; combinations $C(n, r) = \frac{n!}{r!(n-r)!}$ count unordered selections. So picking $3$ from $5$ in order gives $60$ ways; in any order, $10$.

Slide $n$ and $r$. The lab returns $P(n, r)$, $C(n, r)$ and the ratio $P/C = r!$ — the number of orderings of $r$ items. Try $n = 6$, $r = 2$: $P = 30$, $C = 15$, ratio $2! = 2$.

Number of $4$-letter passwords from $26$ letters with no repeats: permutations, $P(26, 4)$. Number of $4$-member committees from $26$ candidates: combinations, $C(26, 4)$. Reading the problem carefully — does order matter? — decides which formula to use.

By convention and to make formulas like $C(n, n) = \frac{n!}{n!\,0!} = 1$ consistent. There is exactly one way to arrange zero objects: do nothing. Setting $0! = 1$ keeps $C(n, 0) = 1$ and the binomial theorem clean. The lab uses this convention silently — try $r = 0$.