Solution:

We count based on how many $1$'s the sequence contains.

If $a = b = c = d = e = 1$ then this gives us $1$ possibility.

If $a = b = c = d = 1$ and $e \neq 1$, $e$ can be $2, 3, 4, 5, 6$. Each such sequence $(1,1,1,1,e)$ can be arranged in $5$ different ways, for a total of $5 \cdot 5 = 25$ ways in this case.

If three of the numbers are $1$, the last two can be $(2,2), (3,3), (2,3), (2,4)$, or $(2,5)$. Counting ordering, this gives a total of $2 \cdot 10 + 3 \cdot 20 = 80$ possibilities.

If two of the numbers are $1$, the other three must be equal to $2$ for the product to be under $10$, and this yields $10$ more possibilities.

Thus there are $1 + 25 + 80 + 10 = 116$ such sequences.