Determine the Probability of no bin being empty when 10 balls are randomly tossed with equal probability into 6 bins Suppose we have 10 items that we will randomly place into 6 bins, each with equal probability. I want you to determine the probability that we will do this in such a way that no bin is empty. For the analytical solution, you might find it easiest to think of the problem in terms of six events Ai, i = 1, . . . , 6 where Ai represents the event that bin i is empty, and calculate P(AC1 ∩···∩AC6)using DeMorgan’s law.
Analytical= 1- P(bin one empty) + P(bin 2 empty) + P(bin 3 empty) + P(bin 4 empty) + P(bin 5 empty) + P(bin 6 empty) – P(1 and 2 empty) – P(1 and 3 empty) – P(1 and 4 empty) – P(1 and 5 empty) – P(1 and 6 empty) – P(2 and 3 empty) – P(2 and 4 empty) – P(2 and 5 empty) – P(2 and 6 empty) – P(3 and 4 empty) – P(3 and 5 empty) – P(3 and 6 empty) – P(4 and 5 empty) – P(4 and 6 empty) – P(5 and 6 empty) + P( 1 2 3 empty) + P(1 2 4 empty) + P (1 2 5) empty + P (1 2 6 empty) + P( 1 3 4 empty) + P(1 3 5 empty) + P( 1 3 6 empty) + P(1 4 5 empty) + P(1 4 6 empty) + P(1 5 6 empty) – P(1 2 3 4 empty) – P(1235 empty) – P(1236 empty) – P(2345 empty) – P(2346 empty) – P(3456 empty) + P(12345 empty) + P(23456 empty) + P(34561 empty) + P(45612 empty) – P(123456 empty)
1-(6P(1 specific empty bin) – 15P(2 specific empty bins) + 20P(3 specific empty bins) – 15P(4 specific empty bins) +6*P(5 specific empty bins)
This is what I have so far but I don't know how to determine the individual probabilities above.
 A: It can be written as,
$\displaystyle \small 1 - 6 \cdot \bigg(\frac{5}{6}\bigg)^{10} + 15 \cdot \bigg(\frac{4}{6}\bigg)^{10} - 20 \cdot \bigg(\frac{3}{6}\bigg)^{10} + 15 \cdot \bigg(\frac{2}{6}\bigg)^{10} - 6 \cdot \bigg(\frac{1}{6}\bigg)^{10}$
$\displaystyle \small \frac{5}{6}$ represents the probability that an item will land into one of the five specific bins out of six (i.e. a specific bin is empty). Similarly $\displaystyle \small \frac{4}{6}$ represents the probability that an item will land into one of the four specific bins and so on.
A: Another way is to use  Stirling numbers of the second kind which can be conceived  as putting n labelled objects (throws $1 \;thru\; 10$)
into k identical boxes with no box being left empty.
If we multiply $S2(10,6)$ by $6!$ to restore the identity of boxes, we immediately get the answer as $\dfrac{S2(10,6)\cdot 6!}{6^{10}} \approx 0.2718$

Addendum
@user2661923 has added a method for manually
working out S2(10,6). Here is another way  which is totally mechanical, without bothering about double counting. There are $5$ possible templates, viz $511111\;421111\;331111\;322111\;222211$ and the answer is
$10!\left[\dfrac 1 {5!\;5!} + \dfrac 1 {4!2!\;4!}+\dfrac 1 {3!3!\;2!4!}+\dfrac1 {3!2!2!\;2!3!}+\dfrac1 {2!2!2!2!\;4!2!}\right]$
To explain the third template say, to get the arrangements, you want $\binom{10}{3,3,1,1,1,1}\binom{6!}{2!4!}$ which simplifies to $\frac{10!6!}{3!3!1!1!1!1!2!4!}$ but we need to eliminate the $6!$ from the numerator and don't need to pad with $1!$
