Given 5 balls, how many ways are there to place them in 3 boxes?

(There are no limitations, i.e. one box can contain all 5, while the rest can contain none.)

Letting the number of balls be n, and the number of boxes be r,

we know the formula is $\displaystyle {{n+r-1} \choose n}$.

Let's call this K(0).

Evaluating, the answer is $\displaystyle {7 \choose 5} \ = \ 21$

If we now include a limitation, such that there be at least 1 ball in each box,

the formula is now altered to:

$\displaystyle {{n-1} \choose {r-1}}$.

Let's call this K(1).

Now comes the question:

The Probability of no empty box is simply $\displaystyle {K(1) \over K(0)}$.

TRUE or FALSE, anyone?