Solution:

Factoring, we get $26 =$ twelve $+$ eleven $+$ two $+$ one $= (elve + o)(tw + ne)$.

Both factors are at least $1 + 1 = 2$, so there are two cases: either $elve + o = 2$ and $tw + ne = 13$, or $elve + o = 13$ and $tw + ne = 2$.

In the first case, $elve = o = 1$, so $e = l = v = o = 1$. We are then left with $tw + ne = 13$.

In the second case, $tw = ne = 1$, so $t = w = e = n = 1$. We are then left with $lv + o = 13$.

Each case has the same number of solutions as $ab + c = 13$. For each $a$, there are $\left\lfloor \frac{12}{a} \right\rfloor$ multiples of $a$ under $13$. Thus the number of solutions to this equation is
$$\sum_{a=1}^{12} \left\lfloor \frac{12}{a} \right\rfloor = 12 + 6 + 4 + 3 + 2 + 2 + 1 + 1 + 1 + 1 + 1 + 1 = 35$$
so the number of solutions to the original problem is $2 \cdot 35 = 70$.