We can easlily eliminate option A & D as $\exists$ with $\rightarrow$ and $\forall$ with $\wedge$
now comes the most confusing part option B & C.
Let's take a counter example -
Suppose we have 3 teachers ($x_{1}$, $x_{2}$ & $x_{3}$) & 10 students ( $y_{1}$, $y_{2}$, $y_{3}$, $y_{4}$, $y_{5}$, $y_{6}$, $y_{7}$, $y_{8}$, $y_{9}$, $y_{10}$).
Now in option B outer for loop is x & inner for loop is y i.e for every value of x there will be multiple y. $\exists y$ is fixed in an particular iteration of x, not on all iteration of x.
for all x i.e $x_{1}$ $\rightarrow$ ($y_{2}, y_{3}, y_{4}$) & $x_{2}$ $\rightarrow$ ($y_{7}, y_{8}, y_{9}, y_{10}$) & $x_{3}$ $\rightarrow$ ($y_{1}, y_{2}, y_{5}, y_{6}$)
So, all of this implies Every Teacher is liked by Some students.( key point is this some is not same for all teacher)
Now in option B outer for loop is y & inner for loop is x, i.e -
there exist some student at the beginning of the statement which says that this is talking about some fixed no. of students, Let's say ($y_{1}, y_{2}, y_{3}$) & ($x_{1}, x_{2}, x_{3}$)
now $y_{1}$ $\rightarrow$ $(x_{1}, x_{2}, x_{3})$, $y_{2}\rightarrow$ $(x_{1}, x_{2}, x_{3})$, $y_{3}\rightarrow$ $(x_{1}, x_{2}, x_{3})$
So,this implies Some students like every teacher.
