The total number of numbers lying between $100$ and $1000$ that can be formed with the digits $1, 2, 3, 4, 5$,if the repetition of digits is not allowed and the numbers are divisible by either $3$ or $5$,is:

The value of ${}^6P_4 + 4 \cdot {}^6P_3$ is $.......$ .

For all $n \in N$,the product $(n+24)(n+25)(n+26)(n+27)$ is always divisible by:

$A$ five-digit number divisible by $3$ has to be formed using the numerals $0, 1, 2, 3, 4,$ and $5$ without repetition. The total number of ways in which this can be done is:

For integers $n$ and $r$,let $\binom{n}{r} = \begin{cases} ^{n}C_{r}, & \text{if } n \geq r \geq 0 \\ 0, & \text{otherwise} \end{cases}$. The maximum value of $k$ for which the sum $\sum_{i=0}^{k}\binom{10}{i}\binom{15}{k-i} + \sum_{i=0}^{k+1}\binom{12}{i}\binom{13}{k+1-i}$ exists,is equal to ...... .

The number of all five-letter words (with or without meaning) having at least one repeated letter that can be formed by using the letters of the word $INCONVENIENCE$ is: