Probability of getting all distinct numbers when picking non unique numbers with replacement sorry in advance if my description is unclear but maths really isn't my strong suit
My question: 
How to determine the probability of picking all distinct numbers  when picking, with replacement, n times from a larger set N where the numbers are non unique AND a specific one of them needs to be in the result set.
Example:
My set of numbers and their probability of being picked are as follows
P(1)=1/20
P(2)=1/20
P(3)=2/20
P(4)=2/20
P(5)=2/20
P(6)=4/20
P(7)=4/20
P(8)=4/20    
What is the probability of picking exactly 5 distinct numbers which include number 8?
I've read questions which were simpler versions of this and understand that if they weren't weighted and there wasn't an inclusion constraint, it would be a matter of doing 1 * 19/20 * 18/20 ... 5 times.
So with my 2 additional conditions I tried to write out the permutations to get an idea starting with picking the number 8 which looks like 
1 * 16/20 * 12/20 * 8/20 * 6/20
1 * 16/20 * 12/20 * 10/20 * 8/20
...
Then I figured I'd try to figure out how to calculate all permutations, but this is where I am stuck because I can't work out how the required number 8 fits in
Thanks for any help
 A: One number needs to be $8$. The remaining numbers can have the following probabilities of being picked (in units of $\frac1{20}$) in the following numbers of ways:
\begin{array}{c|l}
4,4,2,2&\binom22\binom32\binom20=3\\
4,4,2,1&\binom22\binom31\binom21=6\\
4,4,1,1&\binom22\binom30\binom22=1\\
4,2,2,2&\binom21\binom33\binom20=2\\
4,2,2,1&\binom21\binom32\binom21=12\\
4,2,1,1&\binom21\binom31\binom22=6\\
2,2,2,1&\binom20\binom33\binom21=2\\
2,2,1,1&\binom20\binom32\binom22=3\\
\end{array}
Each quintuple of numbers can be picked in $5!$ different orders. Thus the probability for picking $5$ distinct numbers of which one is $8$ is
$$
\frac{5!}{20^5}\cdot4\cdot\left(3\cdot4\cdot4\cdot2\cdot2+6\cdot4\cdot4\cdot2\cdot1+1\cdot4\cdot4\cdot1\cdot1+2\cdot4\cdot2\cdot2\cdot2+12\cdot4\cdot2\cdot2\cdot1\\+6\cdot4\cdot2\cdot1\cdot1+2\cdot2\cdot2\cdot2\cdot1+3\cdot2\cdot2\cdot1\cdot1\right)=\frac{549}{5000}=0.1098\;,
$$
where the $4$ in front is for the probability of picking $8$, the first factor in each term is the multiplicity from the table above and the remaining four factors in each term are for the probabilities of picking the remaining four numbers.
