If 24 pieces of sausage are randomly put onto a pizza what is the probability that your slice will have 3 pieces of sausage? This Question is from Statistics the Easy Way, third edition, by Douglas Downing Ph.D and Jeffery Clark, Ph.D
Chapter 6, Question 1.
If 24 pieces of sausage are randomly put onto a pizza that is sliced into 8 pieces (with none of the sausages getting cut), what is the probability that your slice will have 3 pieces of sausage?
The answer given in the back of the book is the following:
$\binom{24}{3}\cdot \left ( \frac{1}{8} \right )^{3}\cdot \left ( \frac{7}{8} \right )^{21} = 0.006623494492036295$
I do not understand how the authors got this answer. As pieces of sausage are not distinguishable there is only one way to select 3 pieces of sausage. Are the sausage pieces first being labeled and then selected in $\binom{24}{3} $ ways?
Isn't this a balls to boxes problem with indistinguishable balls and distinguishable boxes? 
There are 24 balls and 8 boxes. The number of ways of assigning balls to boxes is $\binom{24+8-1}{8}$.
There is only one way of putting 3 indistinguishable balls in my box and then there are $\binom{21+7-1}{7}$ ways of assigning the other balls to boxes.
So the probability that my slice has exactly 3 pieces of sausage on it is 
$$\frac{\binom{21+7-1}{7}}{\binom{24+8-1}{8}} = 0.11256952169076752$$ 
As these answers are clearly different could someone please explain the error in the way I have solved the problem? 
 A: You can't use stars and bars for this.  Your calculation weights all distributions equally, as though it were just as likely that all $24$ sausages are on the first slice as it is that there are $3$ on each slice, which is clearly not the case.  
The book's calculation is made as if the sausages are placed on the pizza after it has been sliced.  Then the are ${24\choose3}$ ways of choosing which $3$ sausages go on your slice, each has a probability of $1/8$ of landing on your slice, and the other $21$ sausages each have a probability of $7/8$ of not landing on your slice.
Rather a silly question, I think, as nobody makes a pizza like this.
ADDENDUM in response to OP's comment:
We aren't counting the different distributions of sausages: Alice got $4$, Bob got $2$, Carol got $3$, etc.  That's what stars and bars would count, and it's not a question of probability, because we're only counting the outcomes, and not considering their likelihood.  If you divide the number of distribution is which Alice gets $3$ sausages by the total number of distributions, and call that the probability, you are treating all distributions as equally likely, and that's not so.
Consider a simpler problem.  A coin is tossed $10$ times.  What is the probability that all tosses come up heads?  Using the approach you used in this problem, you would say, "This is like tossing $10$ indistinguishable balls into $2$ distinct buckets.  There are $11$ distributions, and only $1$ with all heads, so the probability is ${1\over11}.$"
That's obviously wrong.  The probability is ${1\over2^{10}}={1\over1024}.$ 
A: I agree with the left-hand side of the book's answer: ${24 \choose 3}\cdot\left(\frac18\right)^3\cdot \left(\frac78\right)^{21}$
However this evaluates out to $\frac{1130496828904566830168}{4722366482869645213696}\approx .2394$
