If one of the elements of a triplet of <em>A</em> must be <em>e</em>, then you have
![\dbinom 42 = \dfrac{4!}{2!(4-2)!} = 6](https://tex.z-dn.net/?f=%5Cdbinom%2042%20%3D%20%5Cdfrac%7B4%21%7D%7B2%21%284-2%29%21%7D%20%3D%206)
possible choices for the other two elements in the triplet. They are
{<em>a</em>, <em>b</em>, <em>e</em>}
{<em>a</em>, <em>c</em>, <em>e</em>}
{<em>a</em>, <em>d</em>, <em>e</em>}
{<em>b</em>, <em>c</em>, <em>e</em>}
{<em>b</em>, <em>d</em>, <em>e</em>}
{<em>c</em>, <em>d</em>, <em>e</em>}
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
<em>abe</em>, <em>aeb</em>, <em>bae</em>, <em>bea</em>, <em>eab</em>, <em>eba</em>
<em>ace</em>, <em>aec</em>, <em>cae</em>, <em>cea</em>, <em>eac</em>, <em>eca</em>
and so on.