Combination
In short: A combination is a selection of items in which order does not matter, and the number of ways to choose r from n different items is nCr = n!/(r!(n – r)!).
PrecalculusProbabilityS-CP.B.9
What is Combination?
A combination is a group picked without regard to order. Choosing 4 toppings from a pizza menu gives the same pizza whether you name pepperoni first or last, so only the set of toppings counts.
The count is \(\binom{n}{r} = \frac{n!}{r!\,(n – r)!}\). It equals the ordered count \(P(n, r)\) divided by \(r!\), the number of ways to shuffle the chosen items among themselves. These numbers are symmetric, \(\binom{n}{r} = \binom{n}{n – r}\), and they appear in Pascal’s triangle and in the binomial theorem.
How to spot it
- A committee, team, hand of cards, set of toppings or subset is chosen, and roles are not assigned.
- The wording says choose, select or pick, and no ranking is attached to the picks.
- Reordering the chosen items would give the same outcome, as with a lottery ticket.
- A binomial coefficient such as C(12, 4) or a symbol with n over r appears.
The method
- Confirm that order does not matter and that no item is chosen twice.
- Identify \(n\), the pool size, and \(r\), the number chosen.
- Compute \(\frac{n!}{r!(n – r)!}\), cancelling the larger factorial in the denominator first.
- For several independent choices, multiply the combination counts together.
Combination: solved example
Problem
A pizza shop offers 12 toppings. How many pizzas can be made with exactly 4 different toppings?
Solution
Order does not matter, so use \(\binom{12}{4} = \frac{12!}{4!\,8!} = \frac{12\cdot 11\cdot 10\cdot 9}{4\cdot 3\cdot 2\cdot 1}\).
The numerator is 11,880 and the denominator is 24, so the quotient is 495. The shop can make 495 different pizzas.
Try it
Question 1
Compute \(\binom{7}{5}\).
Show the answer
\(\binom{7}{5} = \binom{7}{2} = \frac{7\cdot 6}{2} = 21\).
Question 2
A store has 6 managers and 8 cashiers. How many ways can 2 managers and 3 cashiers be chosen for a panel?
Show the answer
\(\binom{6}{2}\binom{8}{3} = 15 \cdot 56 = 840\).
Combination vs. Permutation
Ask whether a different order counts as a different result. If a podium finish of gold, silver and bronze is wanted, it does, and you use permutations. If only the set of finalists matters, it does not, and you use combinations. Dividing the first count by \(r!\) converts it into the second.
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