For 𝑖 = 1, 2, 3, 4, let $$𝑇_{𝑖}$$ denote the event that the students $$𝑆_{𝑖}$$ and $$𝑆_{𝑖}$$+1 do NOT sit adjacent to each other on the day of the examination. Then, the probability of the event $$𝑇_{1} \cap 𝑇_{2} \cap 𝑇_{3} \cap 𝑇_{4}$$ is

A

$$\frac{1}{15}$$

B

$$\frac{1}{10}$$

C

$$\frac{7}{60}$$

D

$$\frac{1}{5}$$