$4$ 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.
Best Answer
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.
$$\frac67\cdot\frac57\cdot\frac47 = \frac{120}{7^3}$$