Stars and Bars: Why An Alternate Approach Doesn't Work From http://discrete.openmathbooks.org/dmoi2/sec_stars-and-bars.html, suppose you have 7 cookies to distribute to 4 kids. Now, using stars and bars, we arrive at 10C3 (if you're choosing the 3 bars out of the 10 indices for the stars-and-bars string), or 10C7 (if we're instead choosing the stars). My question is why the answer $(4^7)/7!$ would not also be a valid approach? I understand that the two aren't numerically equal, and that the latter is incorrect, but I'm wondering what my error in reasoning is. For instance, if we imagine the kids as the letters A, B, C, and D, and the cookies as a 7-character string, then $4^7$, to me, seems to be the number of ways to assign the kids to each cookie such that the ordering matters (of which cookie gets assigned which kid mattering, and treating the cookies as distinct). Why, however, can we not then simply divide by $7!$, the number of ways to arrange those characters in a 7-character string, to yield our answer (where which cookie gets assigned to which kid not mattering, and thereby also making which cookie is which no longer be distinct)?
Perhaps it is a simple misunderstanding of ordering vs. distinction, but I'd still like to clarify, if possible.
Thanks!
 A: For one thing, $4^7/7!$ is not even a whole number!
The issue is that, for instance, the string $1122334$ has $\frac{7!}{2! 2! 2! 1!}$ reorderings, not $7!$. But the number of reorderings depends on the number of cookies each child gets (e.g., $1111234$ has $\frac{7!}{4!}$ orderings), so you cannot divide $4^7$ by a single number.
A: The stars and bars solution assumes the cookies are indistinguishable, and the only thing that matters is how many cookies each of the kids gets (the kids, however, are distinguishable). Note that if you number the cookies, it would be impossible in this distribution for kid B to receive a lower-numbered cookie than kid A.
Saying $4^7$ implies that the cookies are distinguishable; because it matters which kids is assigned cookie 1. So giving cookie 1 to kid A, and all other cookies to kid B result in a different distribution than giving cookie 2 to kid A and the remaining six cookies to kid B.
Okay, so let's say that you distribute them as if they were distinguishable, and then you want to divide by something to take into account the fact that they aren't. The thing is that $7!$ is not the correct way to do that, because it is not the case that each distribution of cookies is described in $7!$ ways. Again, consider the distribution of cookies in which $1$ cookie goes to kid A, and the remaining $6$ cookies go to kid B. How many times did you produce that particular assignment? Well, you didn't produce it $7!$ times, you only produced it $7$ times: once when kid A got cookie $1$, once when it was cookie $2$, once when it was cookie $3$, etc. But if it is the distribution that gives two cookies to kid A and the remaining five to kid B was produced $\binom{7}{2}$ times, not $7$.
The number of times each distribution is realized will vary depending on what the distribution is. It is not nearly as simply as just dividing by a fixed quantity, because each final distribution comes from a different number of ordered distributions.
That's why stars and bars is the better way of dealing with this problem.
