Answered

A pedlock has a four digit code that includes digits from 0 to 9, inclusive.

What is the probability that the code does not consist of all odd digits of the same digit is not used more than once in the code ??

Verified Answer :

LammettHash

There are \({}_5P_4=5!\dbinom54=\dfrac{5!}{(5-4)!}=120\) ways of making four-digit codes from the five available odd numerals (1, 3, 5, 7, 9) without replacement. Without any restrictions aside from no replacement, there are \({}_{10}P_4=10!\dbinom{10}4=\dfrac{10!}{(10-4)!}=5040\) possible codes that can be made. So the probability of randomly choosing a code made up only odd digits is

\(\dfrac{120}{5040}\)

which means the probability of this not occurring is

\(1-\dfrac{120}{5040}\approx0.976\)

The answer is 4,920 out of 5,040 or 41/42.

Hope it helps!

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