Textbook
For integers $0\le k\le n$, the binomial coefficient is $\binom nk=n!/[k!(n-k)!]$. It counts ways to choose $k$ of $n$ distinct objects when order is ignored. No object is selected more than once.
An ordered selection first gives $n!/(n-k)!$ possibilities. Each set of $k$ chosen objects appears in $k!$ different orders. Dividing by $k!$ removes that repeated counting. For $k=0$ or $k=n$, the answer is 1: there is one way to choose nothing or everything.
The symmetry $\binom nk=\binom n{n-k}$ follows by matching a chosen set with the objects left behind. Every choice of $k$ objects has exactly one complement of size $n-k$, and conversely. This does not allow selecting an object repeatedly; that would change the admissible collections. The formula also ignores distinct roles within the selected group, since dividing out permutations has already removed internal order.
Definitions and further reading: Mathematics for Computer Science.
In plain language
A binomial coefficient counts fixed-size selections of distinct objects when order does not matter. For a team of two chosen from four, Alice–Bob and Bob–Alice describe one pair, so permutations of the chosen people are removed from the count. Assigning captain and deputy makes the roles part of each outcome and changes the selection model. The formula counts membership rather than reading competition rules or judging the typography of the roster.
Example
The example lists pairs of cards A, B, C, and D. Ordering creates 12 possibilities. Matching each pair with its reverse leaves six selections.
Binomial Coefficients: Choosing Without Order
Ordered selection
Four first choices and three second choices give 12.
Step 1 of 3: Ordered selection
Read all steps
Ordered selection
Four first choices and three second choices give 12.
- State
- ABcurrentACcurrentADcurrentBAcurrentBCcurrentBDcurrentCAcurrentCBcurrentCDcurrentDAcurrentDBcurrentDCcurrent
Duplicate counting
AB and BA represent the same selection.
- State
- AB = BAcurrentAC = CAcurrentAD = DAcurrentBC = CBcurrentBD = DBcurrentCD = DCcurrent
Six selections
After dividing by two, each selection is counted once.
- State
- ABcurrentACcurrentADcurrentBCcurrentBDcurrentCDcurrent
The six selections are explicit: AB, AC, AD, BC, BD, and CD. Each occurs exactly twice among the twelve ordered pairs, which makes the division uniform. Pair AA never appears because the first selected card is unavailable for the second choice. Including it would create another problem; dividing by the old number of permutations could no longer repair the starting list.
Related concepts: Factorials.