There are $4$ balls of different colors and $4$ boxes of the same colors. In how many ways can the $4$ balls be placed in the boxes,one in each box,such that no ball goes into a box of its own color?

$A$ 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 : $D_n = n! \left( \sum_{i=0}^n \frac{(-1)^i}{i!} \right)$

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

Among the $5$ married couples, if the names of $5$ men are matched with the names of their wives randomly, then the probability that no man is matched with the name of his wife is

Three letters are placed at random into three envelopes addressed to three different people. The probability that all letters are placed in the correct envelopes is ..........