How many lineups of 20 are possible where Sally is first, second or third, and Adam is somewhere in the line? The line of 20 is created from 300 students. The next part of the question was to find how many ways there are where Sally is first, second or third. I did a permutation of 299 choose 19 for the lineup (subtracting Sally from both essentially) and then 3 choose 1 for Sally. With the rule of sum, this totalled to (299!/280!)(3). 
Now, I am stuck on how to make Adam in the line too. I was thinking the same idea and take (298!/279)(3)(19) where the 19 represents the ways Adam could be in the line. Am I on the right track here?
 A: Independent of the position of Sally, Adam has $19$ possible positions. So, the total number of possibilities is 
$$3\times 19\times \frac{298!}{280!}$$
A: You are almost right. First place Sally ($3$ options) then place Adam ($19$ options). 
Now $18$ open places are left to be filled. For the first you have $298$ options, for the second $297$ etc.
This results in $$3\times19\times298\times\cdots\times281$$
possibilities.
A: Since both Adam and Sally are in the line, we must choose $18$ of the remaining $298$ people to be in the line.  We have three ways of placing Sally so that she is in one of the first three places.  Once Sally's position has been chosen, we can arrange the other $19$ people (including Adam) in $19!$ ways.  Thus, the number of possible arrangements is $$\binom{298}{18} \cdot 3 \cdot 19!$$
A: First, we pick the $20$ students. One must be Sally and another must be Adam. We choose $18$ others from the $198$ remaining, as you have, to get $\dbinom{198}{18}.$ Now, Sally can have $3$ possible locations, and Adam can have any position from the remaining $19.$ Our answer is $\boxed{\dbinom{298}{18} \cdot 3 \cdot 19!}.$ Adam's presence does not affect the second factor.
