Trying to answer this counting question In a ballroom dance class, participants are divided into couples for each drill session. One partner leads and the other follows for three minutes, and then the couple switches roles for the next three minutes.

(a) Only four people show up on time. How many ways are there to pair
  them up?

My answer here is $4C2 = \binom42 = \tfrac{4!}{2!2!}$

(b) If instead six people show up on time, how many ways are there to
  pair them up?

My answer here is $6C2 = \binom62 = \tfrac{6!}{4!2!}$

(c) Assume all m people in the class arrive on time. (There are an
  even number of people in the class.) How many ways are there to pair
  them up?

My answer here is $mC2 = \binom m2= \tfrac{m!}{(m-2)!2!}$

(d) Consider this time assuming that we specify which member of each
  couple leads first. How many ways are there to pair-and- specify the
  dancers

My answer here is $mP2 = \tfrac{m!}{(m-2)!}$
Are my answers correct?
UPDATE:
Part (a) the answer is $\tfrac{1}{2}(4C2 * 2C2) = 3$
Part (b) the answer is $\tfrac{1}{2}(6C2 * 4C2 * 2C2) = 45$
 A: No, these are not quite correct—at least not the way I understand the question. When there are four people, for example, there are only three ways to pair them up: $\{AB,CD\}, \{AC,BD\}$, and $\{AD,BC\}$. The reason this is different from your answer is that you have counted how many ways there are to choose two people from a set of four. This is ${4 \choose 2} = 6$, namely, $AB, AC, AD, BC, BD, CD$. But the number of ways to form one pair isn't the same as the number of ways to split the group into pairs.
One way to do the count for part (a) is to pick one pair (as you have done), then to form a second pair from the remaining two people. The number of ways to do this is
$${4 \choose 2}{2 \choose 2} = 6 \cdot 1 = 6.$$
But in doing so we have accounted for each pairing twice; e.g., we have counted both $\{AB,CD\}$ and $\{CD,AB\}$, when these are in fact the same pairing. Therefore we should divide by $2$ to get $\frac62 = 3$ pairings. 
A: If I am not mistaken this is applications of the Multinomial Theorem. A lot of good examples are given in Sheldon Ross's A First Course in Probability. 
Also note with part $(d)$ that uniqueness matters. So if we had 4 people show up and we chose who was going to lead who we would have a permutation for all of the members dancing. For example, let the 4 people be named: John, Sally, Jordan, Steve. If John lead Steve for the first 3 minutes it would not be the same as Steve leading John. So essentially we would have the same combination below, but without the denominator in the Multinomial, since order matters. 
A: You already know the answer for part (a).
For part (b), the number of ways to pick a first pair, then a second pair,
then a third pair, is $\binom 62 \binom 42 \binom 22$,
or in your notation $6C2 \times 4C2 \times 2C2$.
But that not only pairs up all the dancers but distinguishes
$\{AB, CD, EF\}$ from $\{AB, EF, CD\}$, $\{CD, AB, EF\}$, and several
other permutations of the pairs.
In fact you have counted the same three pairs $3!$ times, which is
the number of permutations of three things.
So the correct answer is not to divide by $2$ but rather to divide by $3! = 6$:
$$\frac{\binom 62 \binom 42 \binom 22}{6}.$$
Part (d) seems easier to me than part (c).
First you choose who will be leaders in the first part of the lesson. 
There are $\binom{m}{m/2}$ ways to do this. 
Then you line up these $m/2$ leaders in a line and
assign each of the remaining $m/2$ dancers to one of the leaders.
That is, each ordering of the remaining $m/2$ dancers produces a unique
pairing of followers with leaders.
There are $(m/2)!$ such orderings, so the total number of possible
pairings is ...
Once you have part (d), I would go back to part (c).
Clearly there are more pairings counted in part (d) than in part (c).
How many more?
For each way you can pair up dancers in part (c), within each pair
there are two ways to choose who will lead during the first part of the lesson.
Since there are $m/2$ pairs, there are therefore $2^{m/2}$ ways
to choose which $m/2$ dancers will lead at first.
But the choice of pairs for part (c), followed by
choosing leaders for the first three minutes, can give us
every choice of ordered pairs that exists in part (d).
Therefore the number of choices in part (c) must be ...
