Given the set $X=\{x_1,x_2,x_3,...x_{10}\}$ find the number of antisymmetric relations in the set if $x_3\mathrel{R}x_7$, $x_7 \mathrel{R} x_9$ but $x_9 \not \mathrel{R} x_1$.
I wasn't sure if i got this right so i wanted to check. The total number of antisymmetric relation in a set with $n$ elements is calculated to be $2^n3^{\frac{n^2-n}{2}}$. If $x_3\mathrel{R}x_7$, then we are left with only one choice for $x_7$, to not be in relation with $x_1$, so we lose 3 possibilities that we had in the first place. The same thing is true for $x_7 \mathrel{R} x_9$, but for $x_9 \not \mathrel{R} x_1$, we have a possibility that $x_1$ is in relation with $x_9$ and that $x_1$ isn't in relation to $x_9$, so we have 2 possibilities instead of 3 on that field.
The total would then be: $2^{n+1}3^{\frac{n^2-n}{2}-3}$ which would make the answer for my problem $2^{11}3^{42}$.
Is this correct?
Your answer is correct; good job!
Mostly just posting this to get this out of the unanswered queue. Posting as Community Wiki in particular since I have nothing further to add.