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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.

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Depends. We allowing leading zeros? – Neil May 7 '12 at 9:55
No..It must be a 4 digit.No leading zeros. – vikiiii May 7 '12 at 9:57
This changes the problem then. – Neil May 7 '12 at 10:08

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 $A<B<C<D$, 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 $A<B<C<D$. How many four-element subsets are there of a set of ten things?

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This is restating the question. – Neil May 7 '12 at 10:09
@Neil: No, it's providing a very large hint by doing most of the problem and leaving the last step or to for vikiiii to finish. – Brian M. Scott May 7 '12 at 10:10
@Neil: No. $ $ $ $ – Did May 7 '12 at 10:43

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