'c' is a mathematical binary infix operator. (`n` c `r`) is the number of sets with `r` elements which can be made with `n` atoms. (It can also be written C(`n`, `r`), or _{n}C_{r}, or even ^{n}C_{r}.)

Both `n` and `r` must be nonnegative integers, and `r` must be less than or equal to `n`. For example, (4 c 2) is the number of sets with 2 elements which can be made from 4 atoms. Say the 4 atoms are `w`, `x`, `y`, and `z`. The possible sets are: {`w`, `x`}, {`w`, `y`}, {`w`, `z`}, {`x`, `y`}, {`x`, `z`}, and {`y`, `z`}; (4 c 2) is 6. (`n` c `r`) equals (`n`! / ((`n` - `r`)! * `r`!)), which equals ((`n` p `r`) / `r`!).

(`n` c `r`) equals the (`r` + 1)th number on the (`n` + 1)th line of Pascal's triangle, which has many interesting properties. For example, in Eindhoven notation, (+ : (`i` ∈ **Z**) ∧ (0 ≤ `i` ≤ `n`) : `n` c `i`) = 2^{n}.

Phssthpok says the 'c' stands for 'choose'… if you have `n` items available and choose `r` of them, the number of choices available is (`n` c `r`) and it reads as '`n` choose `r`'.

N-wing tells me maybe the infix 'c' should be capitalized, (`n` C `r`). He also says it stands for 'combination', like p (or P) stands for 'permutation'.