$4$ dogs have been born in the same week. What is the probability that they were born on different days?

The first dog can be born on any day.

The second dog has probability 6/7 of being born on a different day.

The third dog has probability 5/7 of being born on a different day.

The fourth dog has probability 4/7 of being born on a different day.

$$\frac67\cdot\frac57\cdot\frac47 = \frac{120}{7^3}$$

There is not enough information to solve this problem. If the dogs were from the same litter the probability is very low that they were born on different days. Normally you should be given some assumption like each day of the week is equally likely (reasonable) and independence (unreasonable - does the problem writer have any idea dogs are usually born into litters?).

In a Bayesian sense learning the four dogs were born in the same week will require updating the prior that their births are not correlated. Given the kennel is not that large.

In short, what a terrible word problem.

The denominator, $7^4$, counts the ways to select a day for each dog. So you must do the same in the numerator: count ways to select a day for each dog (although, distinct days).

You counted ways to select 4 distinct days for the dogs to be born.

However, there are $4!$ ways to assign the dogs to each of these days.

$$\dfrac{{^7\mathsf C_4}\cdot 4!}{7^4} = \dfrac{7!/3!}{7^4} = \dfrac {120}{7^3}$$