So sorry about that @professionalbronco !!

But yes, your solution is absolutely right. Another way of looking at is the "algorithm" Professor Loh mentioned in the video: by sorting alphabetically, you'll get that there are 7 options for person A to make a pair, 5 options for the second pair, 3 for the third, and 1 for the last. $$7\cdot5\cdot3\cdot1=\boxed{105}$$ as well.

Also from the lesson, using the expression you had but expanding out the 4!:
$$\frac{\binom{8}{2}\cdot\binom{6}{2}\cdot\binom{4}{2}\cdot\binom{2}{2}}{4\cdot3\cdot2\cdot1}$$
and then expanding the binomial coefficients and cancelling:
$$\frac{\frac{8\cdot7}{2}\cdot\frac{6\cdot5}{2}\cdot\frac{4\cdot3}{2}\cdot\frac{2\cdot1}{2}}{4\cdot3\cdot2\cdot1} \implies \frac{(\cancel{4}\cdot7)\cdot(\cancel{3}\cdot5)\cdot(\cancel{2}\cdot3)\cdot(\cancel{1}\cdot1)}{\cancel{4}\cdot\cancel{3}\cdot\cancel{2}\cdot\cancel{1}}$$

which leaves us the same expression for \( 7!! \) 🙂

also !!
did you know that the exclamation in the factorial is kind of like the "index" of how much you're shifting each number (double factorial shifts by 2, triple by 3, and so on). Soooo... technically we could get to triple factorials (or higher) maybe:
\(8!!!=8\cdot5\cdot2=80\)
\(11!!!!=11\cdot7\cdot3=231\)

@v4913 I understand now. Thank you very much!

@divinedolphin
Yeah Prof Loh explained in the vid I think.
Also when I was doing the four 4's problem (arrange four 4's and add operations or "smush" - Prof Loh word - them together to get any number from 1-50) I started using !! like mad - around 3/4 of the time (and I was the only person in the class using it frequently)