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

I did:
$$\frac{^7C_4}{7^4}$$
But my book says the solution is:
$\frac{120}{7^3}$
What did I do wrong?
EDIT: I copied the problem exactly as it is in my book. If it is missing information, poorly thought or doesn't make any sense, that's not my fault. Typos, mistakes and low quality abound in these schoolbooks.
 A: You're mistakenly giving an ordering when you use $7^4$. $7^4$ gives the number of $4$-tuples $(w,x,y,z)$ where $1\leq w,x,y,z\leq 7$ with each number representing a day. You get because there are $7$ choices for each place. But in the numerator, you have no ordering- instead, you're counting the number of subsets (and by definition, sets are unordered) of the set $\{1,2...7\}$ with cardinality $4$. As before, each number represents a day, but you're not assigning a day to each dog.
Each set of $4$ days, e.g. $\{1,2,3,4\}$, which we can think of as Sunday, Monday, Tuesday, and Wednesday, represents multiple assignments of days to dogs. The first dog could be born on any of those $4$ days, the second dog could be born on any of those $3$ remaining days, the third dog could be born on any of those $2$ remaining days, and the last dog has no choice. If we multiply the number of sets of size $4$ by $4!=24$, we get all of the permutations of days represented by those sets, since each one has $24$ permutations. This additional factor gives us the given answer.
A: 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.
$$\frac67\cdot\frac57\cdot\frac47 = \frac{120}{7^3}$$
A: 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.
A: 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.
$$\dfrac{{^7\mathsf C_4}\cdot 4!}{7^4} = \dfrac{7!/3!}{7^4} = \dfrac {120}{7^3}$$
A: Obviously, the eldest one has nothing to care about getting birth on a 'different' day, so he/she has $C(7,1)$ choices out of $7$ days of a week with the probability equal to $\frac{C(7,1)}{7}=1$. 
