If one of the elements of a triplet of A must be e, then you have
[tex]\dbinom 42 = \dfrac{4!}{2!(4-2)!} = 6[/tex]
possible choices for the other two elements in the triplet. They are
{a, b, e}
{a, c, e}
{a, d, e}
{b, c, e}
{b, d, e}
{c, d, e}
Multiply this by 3! to account for all possible permutations of a given triplet and you end up with 3! • 6 = 36 possible permutations. They are
abe, aeb, bae, bea, eab, eba
ace, aec, cae, cea, eac, eca
and so on.
