Combinations

Remove the ordering that does not matter

There are ${}_8P_3=336$ ordered ways to select three of eight people. Each unordered group appears in $3!=6$ orders. Divide out those repeated descriptions:

For integers $0\le r\le n$,

The symbol is read “n choose r”. It is not a fraction with a missing line. The expression equals one for $r=0$ and $r=n$: there is one empty subset and one subset containing every object.

The counting-principles reference develops the relationship between ordered arrangements and unordered selections. The division is justified because every selected group has the same number of internal orders.

Choose the right model before applying the formula

Selecting three members for an unranked team uses $\binom n3$. Selecting a first, second and third presenter uses ${}_nP_3$. Choosing three flavours when repeats are allowed is another problem again; the ordinary subset formula does not permit selecting the same object repeatedly.

Selecting r objects is equivalent to deciding which $n-r$ objects to leave out:

For example, selecting six of eight people gives the same 28 possibilities as choosing the two omitted people. This is a counting argument, not just a symmetry visible in the formula.

Impose a membership requirement with disjoint choices

A pool contains five senior members and four junior members. Choose an unranked committee of three containing exactly two seniors and one junior. Choose the two seniors in $\binom52=10$ ways and the junior in $\binom41=4$ ways:

Every valid committee has exactly one senior pair and one junior, so the product counts it once. If all $\binom93=84$ three-member committees are equally likely, the probability of that composition is $40/84=10/21$.

“Exactly two seniors” is essential. A committee containing all seniors does not belong to this event, even though it contains more than one pair of seniors.

Use a complement for at least one member of a category

For the same pool, the number of committees with at least one junior equals all committees minus all-senior committees:

Under uniform selection, the probability is $74/84=37/42$. This is usually shorter than separately counting one, two and three juniors. Those separate cases are valid too, provided each is counted exactly once.

Do not subtract committees with exactly one senior: that is a different event. Translate the requested condition before choosing the complement.

Derive a finite-bag distribution

Choose two tokens without replacement from three A tokens and two B tokens, with every two-token subset equally likely. There are

possible subsets. If X is the number of A tokens selected, then

These probabilities match the earlier ordered-draw tree. The middle probability combines AB and BA; the unordered calculation counts each mixed pair once. Each two-token subset corresponds to exactly two equally likely ordered draws. That constant multiplicity is why the induced subset probabilities are uniform and the two approaches agree.

For integer parameters $0\le K\le N$ and $0\le n\le N$, suppose K of the N objects belong to a designated category. A uniformly chosen n-subset has

for feasible integers $\max(0,n-(N-K))\le k\le\min(n,K)$. This is the hypergeometric model. It is different from binomial sampling because removing an object changes the remaining composition.

Connect combinations to Pascal’s rule

For integers $n\ge2$ and $1\le r\le n-1$, mark one particular object among n objects. An r-subset either contains it or does not. If it contains the marked object, choose the other $r-1$ objects from the remaining $n-1$. If it does not, choose all r from those remaining objects:

For example, $\binom53=\binom42+\binom43=6+4=10$. The two cases are disjoint and exhaustive. This explains the addition pattern in Pascal’s triangle and the coefficients in binomial expansions.

Related topics

Random Variables turns these outcome groups into numerical distributions. Binomial Probability uses combinations when independent trials, rather than a shrinking population, generate the count.