Three digit numbers are formed using the digits $0,1,2,3,4,5$ without repetition of digits. If a numbers is chosen at random, then the probability that the digits either increase (or) decrease is $p$ then the value of $10p$ is 

The total number of possible three digit numbers using the digits $\left\{0,1,2,3,4,5\right\}$ is $n\left(s\right)=5\left(5·4\right)=100$

Let $E$ be the event of getting a number which have digits either increasing order or in decreasing order

The possibilities are 

The number of numbers having 1 in the thousands place and digits in increasing order is $4_{C_2}=6$

The number of numbers having 2 in the thousands place and digits in increasing order is $3_{C_2}=3$

The number of numbers having 3 in the thousands place and digits in increasing order is $2_{C_2}=1$

The number of numbers having 5 in the thousands place and digits in increasing order is $5_{C_2}=10$

The number of numbers having 4 in the thousands place and digits in increasing order is $4_{C_2}=6$

The number of numbers having 3 in the thousands place and digits in increasing order is $3_{C_2}=3$

The number of numbers having 2 in the thousands place and digits in increasing order is $2_{C_2}=1$

$n\left(E\right)=5+3+1+10+6+3+1=30$

Therefore, the required probability  is $\frac{30}{100}=\frac{3}{10}$