Five letters are placed at random in five addressed envelopes. The probability that all the letters are not dispatched in the respective right envelopes is

If there are $6$ letters and $6$ corresponding envelopes,in how many ways can all the letters be placed in the wrong envelopes?

There are five students $S_1, S_2, S_3, S_4$ and $S_5$ in a music class and for them there are five seats $R_1, R_2, R_3, R_4$ and $R_5$ arranged in a row,where initially the seat $R_i$ is allotted to the student $S_i$,$i = 1, 2, 3, 4, 5$. But,on the examination day,the five students are randomly allotted the five seats.
$(1)$ 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
$(A)$ $\frac{1}{15}$ $(B)$ $\frac{1}{10}$ $(C)$ $\frac{7}{60}$ $(D)$ $\frac{1}{5}$

The number of ways in which $4$ letters can be put into $4$ addressed envelopes such that no letter goes into the envelope meant for it is:

$A$ person writes letters to $6$ friends and addresses the corresponding envelopes. In how many ways can the letters be placed in the envelopes so that at least two of them are in the wrong envelopes?
Notation : $D_n = n! \left( \sum_{i=0}^n \frac{(-1)^i}{i!} \right)$

$150$ students took admission. In how many ways can they be divided into three equal sections $A, B,$ and $C$?