Statistical Inference Question From Statistical Inference Second Edition (George Casella, Roger L. Berger)
"My telephone rings 12 times each week, the calls being randomly distributed among the 7 days. What is the probability that I get a least one call each day?"
The answer is .2285, but I don't know how they got it. My reasoning was as follows:
There are 12 calls and thus 13 places to put "day dividers" to produce possible distributions of calls. There should be 6 day dividers for the week. One possibility:
_1|2_3|4_5_6|7|8||9_10_11_12

(1 on Monday, 2 on Tuesday, 3 on Wednesday, 1 on Thursday, 1 on Friday, 0 on Saturday, 4 on Sunday)
There are 13^6 possible distributions (using this method). In order to satisfy 1 call/day, dividers can't be at the beginning or end, nor can they be on top of one another (signifying a day with no calls). This means there are:
11*10*9*8*7*6

Valid distributions and a probability of: 0.069. Where am I going wrong?
(P.S. This isn't homework as I'm not in school.)
Edit: I don't think each "distribution" is equally likely. That's probably my error. But I still don't know how to get to the correct answer :)
 A: Here is an easy approach to solve the problem. Consider the tuples $(n_1,n_2,\ldots,n_7)$ such that $n_i$ are positive integers satisfying $n_1 + n_2 + \cdots + n_7 = 12$. Further, assume for the moment that $n_1 \geq n_2 \geq \cdots \geq n_7$.
You should find that there are only seven such tuples. Take, for example, the tuple $(4,3,1,1,1,1,1)$. The corresponding probability is, according to the multinomial distribution with parameters $n=12$ and $p_1 = p_2 = \cdots = p_7 = 1/7$, 
$$
\frac{{12!}}{{4!3!1!1!1!1!1!}}\bigg(\frac{1}{7}\bigg)^{4}\bigg(\frac{1}{7}\bigg)^{3}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1}\bigg(\frac{1}{7}\bigg)^{1} = 
\frac{{12!}}{{4!3!}}\bigg(\frac{1}{7}\bigg)^{12}.
$$
Now, if we don't assume that $n_1 \geq n_2 \geq \cdots \geq n_7$, then you should multiply the above probability by ${7 \choose 2}2$. Doing the same for the other tuples, you obtain seven probabilities. Their sum is the probability you are looking for. I have done the calculation and the result is $\approx 0.228452440447$.
