Probability of 1 bucket being empty and one bucket having two balls with 7 buckets and 7 differently-colored balls overall Idea is to get a string that is a permutation of $(0211111)$.
I had two approaches, but both seem to miss by two orders of magnitude (expected answer is 32%).
In the first one, I chose 2 balls out of 7 that will end up in the same bucket, and then permute the rest: $$\frac{{{7}\choose{2}}1!+{{7}\choose{5}}5!}{7^7}=0.31\%$$
In the second one, I looked at the number of permutations of string $(0211111)$ as a percentage of all possible string permutations of length 7: $$\frac{3*2*1*1*1*1*1}{7!}=0.12\%$$
Could you please help me find the error in my ways.
 A: Both are incorrect.  You need to learn when to add and when to multiply and what the significance of each action is in counting.  Very loosely speaking, you multiply when you are describing different steps in a larger process of describing an arrangement and you add when you are describing entirely different categories of arrangements.  It is a bit more complicated than that, but you can read more about it elsewhere.
"In the first, I chose 2 balls out of 7 that will end up in the same bucket"  Okay, great start.  But... you forgot to specify which bucket they go into.  "...and then permute the rest"  so you should be multiplying here, not adding.  Also, we already know which balls make up "the rest" so there is no need to use a binomial coefficient here.
Correcting your first approach, we use multiplication principle with the following steps:


*

*Decide which two balls share a bucket.  $\binom{7}{2}$

*Decide which bucket those two balls go into.  $7$

*Decide which bucket will remain empty.  $6$

*From the smallest numbered remaining ball to the largest, in sequence decide which of the remaining buckets each ball goes into.  $5!$
We divide the result by $7^7$, the number of different ways in which we could distribute the balls without restriction.  Notice, these $7^7$ different possibilities are equally likely to occur, and further we cover all possible outcomes.  $7!$ is the incorrect denominator here since $7!$ only describes those arrangements of balls where exactly one ball goes into each bucket and so misses many of the possibilities.
Completing the computations we get
$$\dfrac{\binom{7}{2}\cdot 7\cdot 6!}{7^7}\approx 0.1285$$
I cannot think of an interpretation of the question which gives an answer of $0.32$, so I expect that you either wrote the problem incorrectly, read the answer incorrectly, or there was a typo somewhere.
As for your second approach, it is entirely wrong and I cannot even understand how you came up with your numerator in order to tell you how to correct it.
A: You want the probability that one from seven buckets will be empty, one from the other six buckets will receive two from seven distinct balls, and remaining five balls will be arranged among the rest of the buckets, when each distinct ball may be assigned any one from seven buckets.
$$\dfrac{\dbinom 71\dbinom 61\dbinom 72 5!}{\dbinom 71^7}$$
A: Let's assume that the buckets are distinguishable.
You can start by choosing which bucket will have $2$ balls and which will be empty.
You'll have $7 \times 6$ ways of choosing those (the remaining $5$ buckets only have $1$ option each, that is to have $1$ ball). So, we can choose the number of ball of the buckets in $7 \times 6 \times 1^5 = 42$ ways.
We are multiplying because we want these choices to happen together (we want one to happen and another to happen and etc). Note that summing the number of events is associated to things like $X$ or $Y$ (and, unless they are disjoint, we may have to subtract the number of cases that are common, i.e., that belong to $X$ and $Y$).
Regarding the placement of the balls, we have to choose $2$ out of the $7$ balls for the first bucket (the one with two balls), and we can do it in $\binom{7}{2} = 21$ ways. Note that it does not matter if, for the first bucket, we choose first the green ball and then the yellow ball, or the other way around.
The second bucket (the empty one) has only $1$ way of dealing with it (that is, to leave it empty).
In the other buckets (with a single ball each), we can place the remaining $5$ balls in $5!$ ways (we can think that we have $5$ options for the third bucket, $4$ remaining options for the fourth buckets and so on). Again, we are multiplying because we are considering $X$ and $Y$ and etc.
As for the total number of cases, you have $7^7$ ways of placing each of the $7$ distinct balls in each bucket ($7$ options for the first ball, and $7$ options for the second ball, and ...), since it may happen that the balls are all on the same bucket, or the balls each in its own bucket or some other distribution.
Putting all of this together, we get:
$$
p = \frac{7 \times 6 \times 1^5 \times \binom{7}{2} \times 1 \times 5!}{7^7} \approx 0.1285
$$

I've edited my answer to remove the wrong part and to provide (below) an explanation why my initial reasoning was wrong. Thanks are due to JMoravitz for his comments.
Initially I assumed that the buckets were indistinguishable. I was aiming to solve this problem in a different way, but I forgot one key aspect of these computations:

If we want to compute probabilities by counting the favourable and the total number of events, the events must be equiprobable.

This means that the probability of having all balls in the same bucket would have to be equal to the probability of, for instance, having each ball in one bucket.
However, this is not true in this case:


*

*The probability of having all balls in the same bucket is:
$$
p = \frac{7}{7^7} \approx 8.5 \times 10^{-6}
$$
since we have $7$ options for the bucket that has all balls and $7^7$ ways of placing the balls into the buckets (as before).

*On the other hand, the probability of having each ball in one bucket is:
$$
p = \frac{7!}{7^7} \approx 6.1 \times 10^{-3}
$$
since we have $7$ options for the first ball, $6$ options (buckets that are still empty) for the second ball, and so on.
This example illustrates that each case of the partition of the seven balls is not equiprobable. So, although the Bell numbers tell us that we have $B_7 = 877$ possible ways of partitioning the balls into the buckets, we should not use this information to compute the probabilities (or, in case we really want to use this information, we should weight the cases according to their probability -- so we would be using the conditional probability, which makes the expressions more complex).
