question involving choosing a limited probability space to make the calculations easier I've bumped into a question that includes the answer, however I can't understand that answer and I would appreciate your help understanding it.
the question goes like this:
"In a musical show, there are 3 artists that take part - Elvis, Gaga and Britney.
There is one song that all of them sing together, Elvis sing one duet with Gaga, and one duet with Britney. Also, Britney sings one solo, Gaga sings two solos, and Elvis 3 solos. the order of songs appearance is random. what would be the probability of number of solos in between the 2 duets?"
The answer takes the probability space to be $\binom{8}{2}$ to choose places for the 2 duets. thus $P(X=0)=\frac{7}{\binom{8}{2}}$, $P(X=1)=\frac{6}{\binom{8}{2}}$ and so on...
I can't understand why its ok to totally ignore the sole song that all the three artists sing together... what I did is similar - the probability space I chose is $\binom{9}{2}*\binom{7}{1}$ to show the amount of possibilities to choose two places for the duets and one place for the song that all the three artists sing together.
Then I get $P(X=0)=\frac{\binom{8}{1}\binom{7}{1}+\binom{7}{1}}{\binom{9}{2}\binom{7}{1}}$ because I choose either two attached places for the two duets and then a place for the song of the three, or I choose a trio in $\binom{7}{1}$ options that goes like {solo, all of the three, solo}.
Help would be appreciated
 A: Try to think what happens when you try to find the probability without the 3-duet song. Does the placement of the 3-duet song changes the number of solo songs in between the duet songs?
The total number of combinations of songs is $9!$, Let's eliminate the 3-duet song. Now we have $8$ spots to fill with $8$ song, hence the number of combinations is $8!$. Let $C_n$ be number of combinations such that n solo songs are between the 2 duet songs. thus the probability is $$\frac{C_n}{8!}$$. Let $D_n$ be the number of solos between the duet songs with the 3-duet song So the probability is $$\frac{D_n}{9!}$$. Now let's add the 3-deut song, We have $C_n$ combinations for the first 8 songs and we have to add somewhere the 3-deut song, and we have 9 options for that. We get that $D_n = C_n \cdot 9$, finally we get that
$$ \frac{D_n}{9!} = \frac{C_n \cdot 9}{9!} = \frac{C_n}{8!}$$
as desired.
EDIT: I am adding the full solution so others could understand as well.
Lets calculate $C_n$. First we need to chose $n$ solo songs from 6 solo's, $\binom{6}{n}$, we have $n!$ combinations for the solo songs, because they are different. Now we need to add $1$ duet from each side , we have $2$ options for that because there are $2$ duets. Now let's treat the $n$ solo's and the 2 duets as one song. we are left with $ 6 - n + 1$ songs, so we can arrange them in $(7-k)!$ different ways. finally
$$ C_n = \binom{6}{n} \cdot n! \cdot 2 \cdot (7-n)! = 6!\cdot 2! \cdot (7-n)$$
So the probability is
$$\Pr(X=n)=\frac{C_n}{8!} = \frac{6!\cdot 2! \cdot (7-n)}{8!} = \frac{7-n}{\frac{8!}{2!6!}} = \frac{7-n}{\binom{8}{2}}$$
