Combination¶

Definition¶

Let $A$ be a finite set with $n$ elements, and let $k\le n$ $n$ . A combination is a selection of $k$ elements from $A$ in which order is not taken into account.

Interpretation¶

A combination records only which objects are chosen, not in which order: choosing $A$ then $B$ gives the same combination as choosing $B$ then $A$ . A $k$ -combination of a set is therefore simply a subset with $k$ elements: rearranging the chosen objects gives no new outcome.

Twelve ordered pairs of four letters grouped in boxes of two, AB with BA and so on, next to the six unordered pairs they collapse to.

The figure counts pairs from the four letters $A,B,C,D$ in two steps. First the order is kept: $4\cdot3=12$ ordered pairs. Then the boxes put each pair together with its reversal, such as $AB$ with $BA$ , and each box becomes one unordered pair. Every unordered pair appears exactly $2!=2$ times among the ordered ones, so there are $\binom42=12/2=6$ of them.

The same argument works in general. There are $n(n-1)\cdots(n-k+1)$ ordered selections of $k$ distinct objects ( $k$ -permutations), and each group of $k$ objects can be ordered in $k!$ ways, so the number of combinations is the binomial coefficient $\binom nk=n!/(k!\,(n-k)!)$ $k)!)$ .

To decide which count applies, ask whether swapping two chosen objects changes the result: for a committee it does not, for a president and a secretary it does. Combinations here use each object at most once; when repetition is allowed, unordered selections are counted with stars and bars. In naive probability, numerator and denominator must be counted the same way, both ordered or both unordered; mixing the two gives a wrong ratio.

Key Results¶

Combinations ignore order, while permutations count order.
The number of $k$ -combinations from $n$ distinct objects is $\binom nk$ .
Combinations are used in sampling without regard to sequence.

Examples¶

Choosing three committee members from twelve people is a combination problem.
A five-card poker hand is counted as a combination because the order of the cards in the hand does not matter.
Selecting two spare parts from a box of ten distinct parts is counted by $\binom{10}{2}$ .

Prerequisites¶

Set

Sources¶

Probability Theory for EOR · Week 1

Aliases: k-combination, unordered selection, k-subset, combination without repetition, unordered sample without replacement

Last updated: 20 October 2026 03:02