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Combination

Combination – definition and example, Precalculus math glossary – Mathovore

Key takeaway: A combination is a selection of r items from n distinct items where order does not matter, counted by C(n, r) = n!/(r!(n − r)!).

Grade 12 – PrecalculusS-CP.B.9Probability and Counting

Definition of Combination

Definition

A combination counts groups, not lineups. Choosing three people for a committee gives the same committee whether you name Ana, Ben and Cai in that order or in the reverse. Because rearranging members creates no new group, each group must be counted exactly once.

The standard method starts with the ordered count and then removes the repeats. Every group of \(r\) items can be written in \(r!\) different orders, so dividing the ordered total by \(r!\) leaves one entry per group. These counts are the entries of Pascal’s triangle, and they supply the coefficients in the binomial theorem.

Formula and notation

The number of ways to choose \(r\) items from \(n\) is

\[C(n,r)=\binom\,{n}{r}=\frac{n!}{r!\,(n-r)!}.\]

It is read “n choose r”. Useful facts: \(\binom\,{n}{0}=1\), \(\binom\,{n}{n}=1\), and \(\binom\,{n}{r}=\binom\,{n}{n-r}\).

choose 3 of A, B, C, D, EABCABDABEACDACEADEBCDBCEBDECDEC(5, 3) = 5! / (3! · 2!) = 120 / 12 = 10 groups
Ten different groups, even though each group of three could be written in 6 orders.

Key facts

  • A combination of r items from n pairs with the leftover n − r items, so the two counts match.
  • Committees, hands of cards, pizza toppings and lottery tickets use this count.
  • The counts for a fixed n add up to 2ⁿ across all values of r.
  • It equals the ordered count divided by r!, which removes duplicate orders.

Combination example

Problem

A robotics club has 10 members. The coach will choose 4 of them to attend a regional meet. How many different groups are possible?

  1. The group has no ranking, so use \(C(10,4)=\frac{10!}{4!\,6!}\).

  2. Cancel the 6! tail: the top keeps four factors, 10 down to 7, giving 5,040. The bottom is \(4!=24\).

  3. Divide: \(5040: 24=210\).

There are 210 possible groups.

Common mistake

Watch out

Valentina stops at 5,040, the product of 10, 9, 8 and 7, and calls it the number of groups.

Why it is wrong

That product counts ordered lineups, where Ana-Ben-Cai-Dan and Dan-Cai-Ben-Ana differ. Every combination of 4 members appears \(4!=24\) times , so divide: \(\frac{5040}{24}=210\).

Related terms and lessons

Browse the glossary: A–Z list · Precalculus vocabulary · Probability and Counting

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