Fair ten-sided die (with numbers {1, 2, ... 10}) rolled five times - probability that you get a strictly increasing sequence of five numbers? My initial instinct was to split the possible configurations of five values into cases; i.e. examine the case in which the sequence of five values starts with a "1", starts with a "2", ... starts with 
"5". 


However, I quickly realized that there were subcases for some of these cases (if my sequence of five values starts with "1", I must think about the cases where it starts with "12", "13", ... "17" and discard the cases where it starts with "18", "19", "1 10", as it is impossible to form a strictly increasing sequence of five numbers with those starting numbers). 


Additionally, even amongst the cases where the sequence starts with "12", "13", ... "17", I must discard the cases that have "129", "139", ... "179". Using this logic, I know I could eventually sum up all the possible configurations of five strictly increasing values and divide that by the total number of outcomes to get the probability; however, this method seems very time-consuming. Is there a better way I could approach this problem?
 A: Another way to view it - Let $A$ be the event that the numbers are strictly increasing, and view it as $ A = B \cap C$ where $B$ is the event that the numbers are distinct and $C$ is the event that the numbers are increasing. 
Then $P(A) = P(B \cap C) = P(B) P(C | B).$ The probability that the numbers are distinct is that the first roll is distinct from the previous 0 rolls, the 2nd roll is distinct from the previous 1 roll, the 3rd roll is distinct from the previous 2 rolls etc, so $$P(B) = \frac{10}{10} \cdot \frac{9}{10} \cdot \frac{8}{10} \cdot \frac{7}{10} \cdot \frac{6}{10}$$
Given that the numbers are distinct, the probability that the $5$ numbers are increasing is $1/5!$ since all $5!$ possible orderings are equally likely. So $P(C|B) = 1/5!$ and therefore $$P(A) = \frac{1}{10^5} \binom{10}{5}$$
A: First you need to know the number of strictly increasing sequences of $5$ numbers. This is the same as the number of subsets of size $5$ in a set with $10$ elements, because every subset can be ordered uniquely in an increasing way. This number is $\binom{10}5$. Then the probability of each of these is $\frac1{10^5}$, so the result is
$$\frac1{10^5}\binom{10}5=0.00252$$
