Tree like structure representation helps best here where we start with top level as A, B, C and then start branching based on what the potential 2nd and 3rd options could be.
In simplified form, we want total number of people who picked B last (3rd).
The cases are A,C,B + C,A,B
Firstly, \(\frac{3}{10} = \frac{9}{30}\) marked A, \(\frac{11}{30}\) marked B, so the rest \(\frac{10}{30}\) marked C as first.
Case A,C,B: \(\frac{2}{3}\) of those who ranked A first, ranked B as 2nd. which means \(\frac{1}{3}\) ranked C as second.
\(\frac{1}{3} * \frac{9}{30} =\frac{3}{30}\)
Naturally, all of those \(\frac{3}{30}\) ranked B as third in this case. Because there is no other option left after marking C as second for them.
Case C,A,B: \(\frac{3}{4}\) of those who ranked C first, ranked B as 2nd which means \(\frac{1}{4}\) ranked A as 2nd
\(\frac{1}{4} * \frac{10}{30} = \frac{10}{120}\)
Here also, all of those \(\frac{10}{120}\) ranked B as third in this case too since no other option is left after marking A as 2nd.
Total = \(\frac{3}{30} + \frac{10}{120 }\)
= \(\frac{12}{120} + \frac{10}{120} \)
= \(\frac{22}{120 }\)
There are total of 120 students \( * \frac{22}{120}\) of them = 22
Ans: Option D