If ABCD be any random 4-digit number in base 11
, what is the probability that A < B < C < D?
How i can achieve this? Thanks in advance.
If ABCD be any random 4-digit number in base 11
, what is the probability that A < B < C < D?
How i can achieve this? Thanks in advance.
The desired probability is just the number of four-digit numbers of the desired type divided by the total number of four-digit numbers.
In base eleven there are eleven possible digits. The first digit of a four-digit number can be any of the ten digits other than $0$, and each of the other three can be any of the $11$ digits, so there are $10\cdot11^3$ possible four-digit numbers.
If $ABCD$ is a four-digit number in which $A, then $\{A,B,C,D\}$ is a four-element subset of the set of $10$ non-zero digits, since none of the digits can be $0$. Conversely, if I take any four distinct non-zero digits, they can be used to form exactly one four-digit number $ABCD$ in which $A. How many four-element subsets are there of a set of ten things?