Combinatorics dance class I have a problem that ím trying to solve and I am not completely sure if my answer is correct, I tried looking for it on the web but i cant find a problem quite like it.
I am translating the question into English, and im doing my best to translate it correct, but im sorry if there are some stupid grammar mistakes. 
„During a dance class 4 pairs (a pair consists of one man and one woman) are chosen from 4 men and 7 women. Romeo and Juliet are students in this class, what is the probability that the two will form a pair.
My answer would be 
$$  \frac{\binom{6}{3}3!}{ \binom{7}{4}4!} $$
My thought process is as follows,
There is only one possibility two choose Rome and Juliet, and 1 Possibility to choose the 3 men, so im not writing them down.
But there are 6C3 possible ways to choose the 3 women left, and 3!  possibilities to assign them to the 3 men.
And there are 7C4 ways to choose the 4 women in total and then 4! Ways to form different pairs.
Is this correct? And if not, where did i make a mistake. 
Many thanks for reading this and helping me out
Ps
I tried writing it down in mathjax but wasn’t very successful with that - sorry
 A: Yes, your reasoning is correct.

Replaying your argument . . .

There are ${\large{\binom{7}{4}}}4!$ possible results for the selection of $4$ dance pairs.

Explanation:


*

*There are ${\large{\binom{7}{4}}}$ possible choices for the $4$ women.$\\[4pt]$

*Once the $4$ women are chosen, there are $4!$ ways to match them with the $4$ men.


There are ${\large{\binom{6}{3}}}3!$ possible results for the selection of $4$ dance pairs such that Romeo is paired with Juliet.

Explanation:


*

*There are ${\large{\binom{6}{3}}}$ possible choices for the $3$ women other than Juliet.$\\[4pt]$

*Once those $3$ women are chosen, there are $3!$ ways to match them with the $3$ men other than Romeo.


Therefore the probability that Romeo is paired with Juliet is
$$\frac{\binom{6}{3}3!}{\binom{7}{4}4!}=\frac{1}{7}$$

So that was your argument, and it's fine, but the answer can be found in a much simpler way . . .

After all, Romeo will be paired with some woman, and there is no bias, hence, since there are $7$ women, the probability that Romeo is paired with Juliet is ${\large{\frac{1}{7}}}$.
