Of all the four-digit positive integers containing only digits from the set {2, 4, 6, 8}, what fraction of them have at least one of their digits repeated? Express your answer as a common fraction.
Hi sri340,
We can ask "how many have no repeated digits" ?
This is 4! as all 4 digits must be different.
There are $\displaystyle 4^4$ 4-digit numbers given the digits may be repeated.
Hence $\displaystyle 4^4-4!$ contain at least one repeated digit.
The fraction is $\displaystyle \frac{4^4-4!}{4^4}=\frac{4(4^3-6)}{4(4^3)}=\frac{4^3-6}{4^3}$