Counting & Probability

If one choice has $m$ options and a second, independent choice has $n$ options, the pair has $m\cdot n$ options. Overlapping or exclusive choices are handled by splitting into cases and adding.

In plain terms. When decisions don't affect each other, multiply the number of options. When they can't happen together, add.

Example. A menu with $3$ mains and $4$ drinks gives $3\cdot 4 = 12$ meals.

Ordered picks of $k$ from $n$: ${}_nP_k = \dfrac{n!}{(n-k)!}$. Unordered picks: $\dbinom nk = \dfrac{n!}{k!(n-k)!}$.

In plain terms. If rearranging counts as different, use permutations; if it does not, use combinations. The only difference is dividing out the $k!$ orderings.

Example. Seat $3$ of $8$ in a row (order matters): ${}_8P_3 = 8\cdot 7\cdot 6 = 336$. Choose a committee of $3$ (order doesn't): $\dbinom 83 = 56$.

For "at least one," compute (total) $-$ (ways it fails). The complement usually avoids messy casework.

In plain terms. Counting what you don't want and subtracting is often far easier than counting what you do want.

Example. Probability of at least one head in $4$ flips: the only failure is all tails, $\left(\tfrac12\right)^4 = \tfrac1{16}$, so the answer is $1 - \tfrac1{16} = \tfrac{15}{16}$.