Number of ways to pair people but with restrictions I have a question regarding combinatorics that I hope you guys can help me clarify:
There are 14 people. 3 of them are TAs. The professor wants to group them into pairs such that no TAs are paired with each other. Find the number of ways to have such arrangement.
So what I'm thinking is that we can use complementary. Without any restrictions, we can have $\frac{14!}{2^7 7!}$ ways to arrange people.
Now we have to think about how to arrange people such that at least one pair is a TA. Suppose the first three people are TA.
So there are 2 ways to pair the first TA with another one. For the second pair is the 3rd TA, so there are 11 ways to select a non-TA. Continue we have $2\cdot 9\cdot7 ... 1$ ways.
So there are $\frac{14!}{2^7 7!} - 2\cdot 9\cdot7 ... 1$ ways. Is this correct? Thanks a lot!
 A: Method 1:  Line up the TA's in alphabetical order.  Do the same for the students.
There are $11$ ways to select a student to pair with the first TA, $10$ ways to select a student to pair with the second TA, and $9$ ways to select a student to pair with the third TA.  Remove those students from the line. That leaves us with eight students.  There are $7$ ways to pair a student with the first student remaining in the line.  Remove that pair.  There are $5$ ways to pair a student with the first student remaining in the line.  Remove that pair.  There are $3$ ways to pair a student with the first student remaining in the line.  Remove that pair.  The two students remaining in the line form the final pair.  Hence, there are 
$$11 \cdot 10 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$$
ways to group $11$ students and $3$ TA's into pairs so that no two TA's are in the same pair.
Method 2:  We correct your approach.
We subtract the number of pairs in which two TA's form a pair from the number of ways the $14$ people could be grouped into pairs.
To count the number of ways $14$ people could be grouped into pairs, line up the $14$ people in some order, say alphabetically.  There are $13$ ways to match a person with the first person in line.  Remove that pair.  There are $11$ ways to match a person with the first person remaining in line.  Remove that pair.  There are $9$ ways to match a person with the first person remaining in line.  Remove that pair.  Continuing in this way, we see that there are
$$13!! = 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$$
ways to group $14$ people into pairs.
There are $\binom{3}{2}$ ways to select a pair of TA's to be in a group together.  That leaves $12$ people to be grouped into pairs.  Reasoning as above, there are
$$11!! = 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$$
ways to group them into pairs, so there are 
$$\binom{3}{2}11!! = 3 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$$
ways to group the $14$ people into pair in such a way that two of the TA's are in the same pair.
Hence, the number of ways to group $11$ students and $3$ TA's into pairs so that no two TA's are in the same pair is
\begin{align*}
13!! - \binom{3}{2}11! & = 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1 - 3 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1\\
                       & = (13 - 3) \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1\\ 
                       & = 11 \cdot 10 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1
\end{align*}
which agrees with the result we obtained above. 
A: How about the following direct approach?

Let us take the three TAs first, and call them $T_1, T_2, T_3$. Now there are $11$ ways of choosing a person to be included in a pair with $T_1$, and corresponding to each of these there are $10$ ways of choosing a person to be paired with $T_2$, and corresponding to each of these pairings of $T_1$ and $T_2$, there are $9$ ways of choosing a person to be paired with $T_3$, thus giving us $11 \times 10 \times 9 = 990$ distinct ways of pairing people with our TAs.
And, corresponding to each of these $990$ ways of pairing people with our TAs, there are ${ 8 \choose 2} = 28$ ways of chosing the remaining pairs.
Thus in all we have $11 \times 10 \times 9 \times 28 = 990 \times 28= 27720$ distinct pairs in all.

Hope I have been able to get this right?
And, hope my solution is clear enough too.
