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?

The number of all four-digit numbers which do not have four distinct digits is

There are $3$ sections in a question paper and each section contains $5$ questions. $A$ candidate has to answer a total of $5$ questions,choosing at least one question from each section. Then the number of ways,in which the candidate can choose the questions,is

Let $n = 1! + 4! + 7! + \ldots + 400!$. Then the ten's digit of $n$ is

$A$ man $P$ has $7$ friends,$4$ of them are ladies and $3$ are men. His wife $Q$ also has $7$ friends,$3$ of them are ladies and $4$ are men. Assume $P$ and $Q$ have no common friends. Then the total number of ways in which $P$ and $Q$ together can throw a party inviting $3$ ladies and $3$ men,so that $3$ friends of each of $P$ and $Q$ are in this party,is . . . . . . .

The number of different $5$-digit numbers greater than $50000$ that can be formed using the digits $0, 1, 2, 3, 4, 5, 6, 7$,such that the sum of their first and last digits is not more than $8$,is: