How many unique combinations to wear seven different socks? My friend has seven unique but similar socks.

How many ways does he have to wear them, not double counting same combos?

$\mathbf {Attempt}$

I think $7! - 5!$ , but then you forget you can put the first foot sock on the other foot as well so $7!-5!+6$ is the correct answer?
You are choosing $2$ socks from $7$ socks. That is ${7\choose 2}=\frac{7!}{2!5!}=21$, where you can first choose from $7$ socks, then you can choose from the $6$ remaining socks (so $7\cdot 6=\frac{7!}{5!}$) and due to counting each pair double, you divide by $2!=2$.

I think it's pretty well explained here, also for the general case $\frac{n!}{k!(n-k)!}$:
https://betterexplained.com/articles/easy-permutations-and-combinations/
You have $7$ choices for your left root. Then you have $6$ choices for the right. This gives $42$ choices. However, this counts $AB$ and $BA$ as different, while the problem says to not double count the same pair. So we have half that many choices, so $21$ is the answer.