I am working with generalized permutations with problems such as the following:
Compute $P(10;4,4,2)$: Answer by formula: $\dfrac{10!}{(4!\, 4! \,2!)}$
A similar problem asks to Compute $P(17;4,3,2)$. It is my understanding that N ($17$ in this case) is supposed to equal the sum of $n_{1}$, $n_{2}$, and $n_{3}$ ($4$, $3$, and $2$ respectively).
Would it be wrong to compute the answer in the same way? $\dfrac{17!}{(4!\,3!\,2!)}$ ?
You have already got the answer, but I'd like to explain a point, and an approach.
Although I haven't encountered it before, P(10;4,4,2) is just the multinomial coefficient $\binom{10}{4,4,2}$
Now $\binom{10}{4,4,2} = \binom{10}{4}\binom64\binom22 = \frac{10!}{4!4!2!},$ and when expanded as a product of binomial coefficients,
the last term e.g. $\binom22$ here, always evaluates to $1$
So sometimes, the last term in the lower index is omitted, e.g. asked how many ways there are to get four $6's$ and four $1's$ in $10$ throws, we might abbreviate it to $\binom{10}{4,4}$
For your particular question, using your notation,
I would thus prefer to write the answer as $P(17;4,3,2,8)$