A quick doubt. Consider the two following questions:
How many ways are there to put 7 indistinguishable balls into 4 boxes in a row?
How many ways are there to put 7 indistinguishable balls into 4 distinguishable boxes?
While the first is simply $\binom{7+4-1}{7}$, the second I would guess it would be $\binom{7+4-1}{7}\times4!$. However, in here, it says that the answer is still $\binom{7+4-1}{7}$ (Theorem 4).
Which is the right answer? Did I miss something?