Counting methods
Many probabilities require counting outcomes that are too numerous to list. Systematic counting rules make this possible.
The multiplication principle
If one choice can be made in \(m\) ways and a second in \(n\) ways, the pair can be made in \(m \times n\) ways. A meal with 3 starters, 5 mains, and 4 desserts has \(3 \times 5 \times 4 = 60\) combinations. A 4-digit PIN has \(10^4 = 10{,}000\) possibilities.
Factorials
The number of ways to arrange \(n\) distinct objects in order is \(n! = n(n-1)\cdots 2 \cdot 1\). Five books can be arranged on a shelf in \(5! = 120\) ways. By convention \(0! = 1\). Factorials grow extremely fast: \(10! = 3{,}628{,}800\).
n! = n(n-1)\cdots 1Permutations
When order matters, the number of ways to choose \(r\) of \(n\) objects is \(P(n, r) = \frac{n!}{(n-r)!}\). Awarding gold, silver, and bronze among 8 runners: \(8 \cdot 7 \cdot 6 = 336\) ways.
P(n,r) = \frac{n!}{(n-r)!}Combinations
When order does not matter, divide out the \(r!\) orderings: \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\). Choosing a committee of 3 from 8 people: \(\binom{8}{3} = 56\). The key question is always: does the order of selection matter?
\binom{n}{r} = \frac{n!}{r!(n-r)!}Worked example: lottery odds
A lottery draws 6 numbers from 49, order irrelevant. There are \(\binom{49}{6} = 13{,}983{,}816\) possible tickets, so one ticket's chance of winning is about 1 in 14 million. Counting reveals just how unlikely such events are.
Probability with counting
With equally likely outcomes, probability = favourable count / total count. The probability that a 5-card poker hand is all hearts is \(\frac{\binom{13}{5}}{\binom{52}{5}} = \frac{1287}{2{,}598{,}960} \approx 0.0005\).
The birthday problem
In a room of 23 people, the chance that at least two share a birthday is over 50%. Use the complement: the probability all birthdays differ is \(\frac{365}{365}\cdot\frac{364}{365}\cdots\frac{343}{365} \approx 0.49\). Intuition fails because there are \(\binom{23}{2} = 253\) pairs, each a chance for a match.
Common mistakes
Using permutations when order does not matter, or combinations when it does. Forgetting that \(0! = 1\). Double-counting arrangements. Treating 'with replacement' and 'without replacement' as the same.
Key takeaways
Multiply the number of choices at each stage. Arrangements of \(n\) objects: \(n!\). Ordered selections: \(P(n, r)\). Unordered selections: \(\binom{n}{r}\). Counting plus equally likely outcomes gives probabilities, often with surprising results like the birthday problem.
Flashcard
m ways then n ways gives mn ways
Check your understanding
How many ways can a president, vice president, and treasurer be chosen from 10 people?