I have a question that reads:
There are 26 people, and 16 tables each with two seats. All people must sit down, what is the % probability that at least one table will be empty?
There is another way to do this other than the way I am going to ask, but my question is rather WHY my method will not work, rather than the answer to this question.
This is my method: Find out how many instances that there is at least one person at each table, then subtract this from the total number of permutations.
So what I do is:
Pick 16 of the 26 people.
26P16
Put one of each of these guys at each table
16P10 * 32 * 30 * 28 *26 * 24 . . . . . * 2
as when I place someone at the first table, I only have 30 seats to choose from, I can't stick him next to the other guy, I want 1 at each of the 16 tables.
Next, of the 16 seats remaining, pick a random 10 for the last 10 guys to sit at
16P10 * 32 * 30 ... * 2 * 16P10
However after the 2nd stage, ever before multiplying by 16P10, this number is massive compared to the total number of permutations. Why am I over counting after the 2nd step?
Cheers for your time.
Consider the multiset of table seats $S=\{s_1,s_1,s_2,s_2,\dots,s_{16},s_{16}\}$, where $s_i$ is a seat in the $i$-th table. Consider the multinomial coefficient
$$ \pmatrix{n\\m_1,\dotsc,m_k}=\frac{n!}{m_1!\cdots m_k!}\,, $$
where $\sum_{j=1}^km_k=n$.
It counters the number of ways of assigning each of $n$ objects to one of $k$ containers, where exactly $m_j$ objects are assigned to the $j$-th container. In our case, the objects are people $(n=26)$, the containers are tables $(k=16)$ and at most $2$ people may be assigned to each container $(m_j\leq 2$ for all $j)$.
Hence, the total number of ways of seating $26$ people round the $16$ tables is
$$T= \sum_{\substack{\displaystyle m_1+\,\dots\,+m_{16}=26\\\displaystyle0\leq m_1,\dots,m_{16}\leq 2}}\pmatrix{26\\m_1,\dotsc,m_{16}}$$
Now, we want to consider the number $E$ of seatings that leave at least one table empty. Instead, let's calculate the number $N$ of seatings that leave no table empty. We will then obtain $E$ from $T=E+N$.
To leave no table empty, it suffices to modify the indices of the last sum to take into account only cases when all $m_j \geq 1$, that is, at least one person is assigned to each container. Thus:
$$N= \sum_{\substack{\displaystyle m_1+\,\dots\,+m_{16}=26\\\displaystyle 1\leq m_1,\dots,m_{16}\leq 2}}\pmatrix{26\\m_1,\dotsc,m_{16}}$$
Finally, the desired probability will be given by $E/T = (T-N)/T=1-N/T$.
EDIT: Let's elaborate on the answer with a more direct calculation. We can calculate $N$ as follows: let $a$ be the number of $m_j$'s that are $1$'s and let $b$ be the number of $m_j$'s that are $2$'s. Then
$$\left\{ \begin{align} &a+b=16\\ &a+2b=26 \end{align} \right.$$
so that $a=6$ and $b=10$. Hence, there are $16\choose6$ terms in the sum expression for $N$ above, each of which equals $\frac{26!}{{(1!)}^6 {(2!)}^{10}}$. It follows that
$$N= \binom{16}{6} \cdot \frac{26!}{2^{10}}=3153865254591658134528000000$$
For $T$, we adapt the calculation slightly. First, for each possible value of $k$, choose $k$ of the $m_j$'s to be $0$, in $16\choose k$ ways. Notice that $0 \leq k \leq 3$. Then, using the same notation as before, we'll have that
$$\left\{ \begin{align} &a+b=16-k\\ &a+2b=26 \end{align} \right.$$
which implies $a=6-2k$ and $b=10+k$. Now the reasoning is as before: for each of the remaining $(16-k)$ $m_j$'s (the ones that are nonzero), $(6-2k)$ of them must be $1$'s and the remaining $(10+k)$ must be $2$'s, so that
\begin{align} T &= \sum_{k=0}^3 \binom{16}{k}\binom{16-k}{6-2k}\frac{26!}{2^{10+k}}=\frac{16!\,26!}{2^{10}}\cdot\sum_{k=0}^3\frac{1}{k!\,(6-2k)!\,(10+k)!\,2^k}\\ &=8557340690780163330048000000 \end{align}
This yields $N/T=143/388$, so that the desired probability is $245/388 \simeq 63.14\%$. Honestly I find this surprisingly high.