Number of ways to divide a group of N people into 2 groups I've seen a bunch of questions about dividing a group of $N$ into groups of a specified size, but I am unsure about how to calculate the total number of ways to divide a group of $N$ people into $2$ distinct groups..
The questions states that one group could be empty, and that a group could have sizes from $0, 1, 2, ..., N$. 
The question then goes on to ask what is the probability that one of the groups has exactly $3$ people in it. I presume this would be calculated by dividing $N\choose 3$ by the total number of ways calculated above, but any other comments would be greatly appreciated!
 A: Suppose you lined every one of them up, and you could assign everyone a $0$ or $1$, for either group. Then each person could either have a $0$ or $1$. So for $N$ people, there are $2^N$ ways of doing this so $2^N$ different groups could be formed. Now to find the probability that one of these groups is of size $3$, how many ways can you pick $3$ people from $N$? Knowing this, you can calculate the probability.
A: The wording of the problem is
"to calculate the total number of ways to divide a group of N people into 2 distinct groups.."
which means two labelled groups.
Since each individual can go to either of the groups (Tigers or Lions, say)
number of possible groups = $2^N$
$\binom{N}{3}$ counts the number of ways Tigers, say, has $3$ members, and another $\binom{N}{3}$ counts similarly for the Lions, except for the special case when $N=6$, where counting a group of $3$ for the Tigers automatically yields a group of $3$ for the Lions, thus
$$Pr = 
\begin{cases}
\dfrac{2\binom{N}{3}}{2^{N}} && \text{if $N \neq 6$}\\[2mm]
\dfrac{\binom{6}{3}}{2^6} && \text{if $N = 6$}
\end{cases} 
$$  
A: If you have $N$ persons, choose any subset of them, form a group with them, and form another group with the rest of them. Since there are $2^N$ subsets of a set with $N$ elements, there are $2^N$ ways of dividing the persons into two groups.
While doing this, there are $\binom n3$ ways of choosing a group with exactly $3$ persons. So, the probability that the first group has $3$ persons is $2^{-n}\binom n3$ and therefore the probability that one of the two groups has $3$ persons is twice that, that is, $2^{-(n-1)}\binom n3$, unless $n=6$. In that case, one of the groups has $3$ persons if and only if the other group has three persons. So, the answer is $2^{-6}\binom63$.
A: Clearly, the probability is $0$ when $N < 3$.
We can count the number of ways of dividing a nonempty set of $N$ elements into two groups in two ways.


*

*We choose a subset, placing the remaining elements of the set in its complement.  We have two choices for each of the $N$ elements, to include it in the subset or not to include it.  Hence, there are $2^N$ subsets.  However, we have counted each choice twice, once when we choose a subset and once when we choose its complement.  Hence, the number of ways of dividing a nonempty set with $N$ elements into two groups is $$\frac{2^N}{2} = 2^{N - 1}$$

*Suppose $x$ is a particular element of the set.  The two groups are completely determined by choosing which of the remaining $N - 1$ elements are in the same subset as $x$.  There are $2^{N - 1}$ ways to choose a subset of the remaining $N - 1$ elements, so there are $2^{N - 1}$ ways of dividing the set into two groups.


As a check, consider the set $\{a, b, c\}$.  The ways we can divide it into two groups are:
$$\emptyset, \{1, 2, 3\}$$
$$\{1\}, \{2, 3\}$$
$$\{2\}, \{1, 3\}$$
$$\{3\}, \{1, 2\}$$
There are $2^{3 - 1} = 2^2 = 4$ ways to divide the set into two groups, as we would expect.
Thus, at first glance, the probability that one of the groups has exactly three people in it is 
$$\frac{\binom{N}{3}}{2^N}$$
However, the case $N = 6$ is special.  When a subset of three people is selected from a group of six people, its complement also has three people.  Hence, choosing a subset of three people counts each way of dividing the group of six people into three people twice, once when we select that subset and once when we choose its complement.  Hence, there are 
$$\frac{1}{2}\binom{6}{3}$$ 
ways to divide a group of six people into two groups of three people.
Alternatively, if Andrew is one the six people, there are $\binom{5}{2}$ ways to select the other two people in his group of three.  Hence, the probability of choosing a subset of three people when a group of six people is divided into two groups is 
$$\frac{\frac{1}{2} \cdot \binom{6}{3}}{2^5} = \frac{\binom{5}{2}}{2^5}$$
Hence, the desired probability is 
$$P = 
\begin{cases}
\dfrac{\binom{N}{3}}{2^{N - 1}} && \text{if $N \neq 6$}\\[2mm]
\dfrac{\binom{5}{2}}{2^5} && \text{if $N = 6$}
\end{cases} 
$$
where we note that $\binom{N}{3} = 0$ when $N < 3$.
