The number of ways of dividing $52$ cards amongst four players equally is

Let $A = \{x_1, x_2, x_3, x_4\}$ and $B = \{y_1, y_2, y_3, y_4\}$. $A$ function $f: A \to B$ is defined. The number of one-one functions such that $f(x_i) \neq y_i$ for $i = 1, 2, 3, 4$ is equal to:

The set $S = \{1, 2, 3, \dots, 12\}$ is to be partitioned into three sets $A, B, C$ of equal size such that $A \cup B \cup C = S$ and $A \cap B = B \cap C = C \cap A = \emptyset$. The number of ways to partition $S$ is:

There are $8$ different coloured balls and $8$ bags having the same colours as that of the balls. If one ball is placed at random in each one of the bags,then the probability that $5$ of the balls are placed in the respective coloured bags,is

There are $4$ balls of different colors and $4$ boxes of the same colors as the balls. In how many ways can $2$ balls be placed in the boxes such that they do not match their respective colors?

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
$(A)$ $\frac{1}{15}$ $(B)$ $\frac{1}{10}$ $(C)$ $\frac{7}{60}$ $(D)$ $\frac{1}{5}$