The following is essentially the same as the solution of WimC. The language is a little different.
For no good reason, except that I like sequences not to decrease, let us list the results in non-decreasing order.
Consider for example the sequence $2,3,3,4,7,7$. We can describe this sequence in an alternate way. The first number is $1$ from the bottom. The next number is $1$ from the previous one. The next number is $0$ from the previous one. The next number is $1$ from the previous one. The next number is $3$ from the previous one. The next number is $0$ from the previous one. Finally, not necessary, but very useful for the later analysis, the biggest possible number, $8$, is $1$ from the previous one.
Call the sequence $(1,1,0,1,3,0,1)$ the encoding sequence of $2,3,3,4,7,7$.
The encoding sequence has length $7$. This $7$ comes from the fact that the sequence we are encoding has length $6$. A sequence of length $k$ will have encoding sequence of length $k+1$.
The sum of the elements of the encoding sequence is $7$. This $7$ comes from the fact that our numbers come from $1,2,\dots,8$. If it were an $n$-sided die, we would have $n-1$ for the sum of the elements of an encoding sequence.
We want to count the number of encoding sequences of length $k+1=7$ with sum $n-1=7$.
So we want to count the number of ways to distribute $n-1$ identical candies among $k+1$ babies.
This is a standard "Stars and Bars" problem (see Wikipedia if you have not met the problem). But we do the details.
We want in this case to distribute $7$ identical candies among $7$ babies. Sorry again about the fact these numbers happen to be equal. This is a pure accident. Explanation would be clearer if we had a $20$-sided die, and we were counting non-increasing (or non-decreasing) sequences of length $6$.
Back to the babies. Being nasty, I like to distribute candies as follows. I distribute $14$ candies to the babies, at least one to each baby. Then I take away one candy from each baby (easy to do).
For the distribution, line up the candies in a row, with a little gap between them. There are $13$ gaps. Choose $6$ of these gaps, because there are $7$ babies. Put a divider into each chosen gap. Give all the candies from the left end up to the first divider to the first baby, the candies between the first two dividers to the second baby, and so on.
The number of ways to distribute the $14$ candies, at least one to each baby, is the number of ways of placing the dividers. This is $\dbinom{13}{6}$.