Five-digit numbers are formed using the digits $1, 2, 3, 4, 5, 6$,and $8$. What is the probability that they have even digits at both the ends?

$A$ certain question paper contains three parts $A, B, C$ with four questions in part $A$,five questions in part $B$,and six questions in part $C$. $A$ student is required to answer seven questions,choosing at least two questions from each part. The total number of different ways a student can choose his seven questions for answering is:

The number of natural numbers lying between $1012$ and $23421$ that can be formed using the digits $2, 3, 4, 5, 6$ (repetition of digits is not allowed) and divisible by $55$ is $....$

If $n$ is an integer with $0 \leq n \leq 11$,then the minimum value of $n!(11-n)!$ is attained when a value of $n$ equals to

$A$ test containing $3$ objective-type questions is conducted in a class. Each question has $4$ options and only one option is the correct answer. No two students of the class have answered identically,and no student has written all correct answers. If every student has attempted all the questions,then the maximum possible number of students who have written the test is:

Let $S$ be the set of all seven-digit numbers that can be formed using the digits $0, 1$ and $2$. For example,$2210222$ is in $S$,but $0210222$ is not in $S$. Then the number of elements $x$ in $S$ such that at least one of the digits $0$ and $1$ appears exactly twice in $x$,is equal to $....$