Size of a type class ($|T_n(q)|=n!/{\prod }_a(nq(a))!$)

Write $n_a=nq(a)$. Arranging $n_a$ copies of each symbol gives the multinomial coefficientThe division by the factorials removes permutations among identical copies. In the balanced four-symbol example, $|T_4(q)|=4!/(2!2!)=6$. For binary symbols this becomes the binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$; for three symbols with counts $(2,1,1)$, the class contains $4!/(2!1!1!)=12$ strings.