Choosing $2$ of the $3$ letters $A$, $B$, and $C$

Terms

Term	This example	Why
number of permutations	$P(3, 2) = 6$	Choosing $2$ of the letters $A$, $B$, $C$ in order gives $P(3, 2) = 3 \cdot 2 = 6$ permutations, because there are $3$ choices for the first letter and then $2$ for the second. The six permutations are $AB$, $BA$, $AC$, $CA$, $BC$, and $CB$.
binomial coefficient	$\binom{3}{2} = 3$	Choosing $2$ of the letters $A$, $B$, $C$ without order gives $\binom{3}{2} = 3$ combinations, because each combination, such as $\setdef{A, B}$, comes from $2! = 2$ permutations, such as $AB$ and $BA$. The three combinations are $\setdef{A, B}$, $\setdef{A, C}$, and $\setdef{B, C}$.

Number of permutations vs. binomial coefficient: Choosing $2$ of the letters $A$, $B$, $C$ gives twice as many permutations as combinations, because each combination of $2$ letters can be put in order in $2! = 2$ ways.