Counting children as unordered binary sequences I am reviewing basic counting techniques and I came across the following problem (in one form or another):
What is the probability that a couple with six children has three boys and three girls? Children are distinguished by gender, assuming each is either a boy or a girl and either outcome is equally likely and independent of the previous outcomes, e.g., one boy and five girls.

My solution is the following. There are six possible unordered 6-sets
  of the letters $B$ and $G$ (fix one value for $B$, this determines
  $G$, etc.), and only one such set has three of each. So the
  probability is 1/6.

This seems correct but surprisingly high a probability. Is it incorrect? And much more importantly for my purposes, can it be improved?
 A: No, you cannot treat this problem like choosing some unordered sequence of $B$'s and $G$'s uniformly at random. You have to treat it as choosing an ordered sequence of $B$'s and $G$'s uniformly at random - in which case there are ${6\choose k}$ sequences having $k$ instances of $B$, out of a total of $2^6$ equally likely sequences.
You should consider what you are given: there are six children, each assigned $B$ or $G$ at random with probability $1/2$ each and independently of previous assignments. The most natural thing to do then is to consider your sample space - the space of possibilities - as the sets of ordered sequences of $6$ terms from $B$ or $G$ - like $BBGBGG$ or $GGGBGB$ and so on. Then, you find that every sequence is equally probable using the givens: this is because, if we build these sequences up one term at a time, we find that starting $B$ is equally likely to starting $G$ and then the next term does not depend on the last one - and so on.
It might help to think about this as a series of coin-flips instead - which models the same probabilities. If you think about the extreme case of flipping the coin $6$ times and getting only heads, your reasoning suggests that you get this $1/6$ of the time, which is way too high, which shows that something is wrong. It's important to remember that probability is not just "count the things, divide by how many things" - you can only do that if you establish that all the things are equally likely, and if you skip this step of justification, you can easily end up with the wrong answer, which is exactly what happened in your solution (since the unordered sequences, while they can model the problem, they are not equally likely). 
A: How many gender combinations in total? Each child has 2 options, so $2^6=64$.
How many ways are there to have exactly 3 boys out of 6 children? This is the same as choosing 3 things from 6, so it's $\binom{6}{3}=20$.
Since all combinations are equally likely, the probability is $\frac{20}{64}=0.3125$.
