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($1$) The probability that, on the examination day, the student $S_1$ gets the previously allotted seat $R_1$, and $NONE$ of the remaining students gets the seat previously allotted to him/her is
$(A)$ $\frac{3}{40}$ $(B)$ $\frac{1}{8}$ $(C)$ $\frac{7}{40}$ $(D)$ $\frac{1}{5}$
($2$) 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
$(A)$ $\frac{1}{15}$ $(B)$ $\frac{1}{10}$ $(C)$ $\frac{7}{60}$ $(D)$ $\frac{1}{5}$
Give the answer or quetion ($1$) and ($2$)