Eight people are to form two queues of four. In how many ways can this be done if: Jim will only stand in the left queue.
I would think you deal with the restriction first so you put Jim into the left queue then permute the rest. So would the answer not be 7P7 = 5040 What am i doing wrong?
As Mark said, you can't just say Jim is in the left queue- he also has a position within that queue.
Another way of looking at it, which is similar to one of your earlier questions. Consider all permutations of $12345678$. We can say that the first four digits is the left queue, in that order, and the last four is the right queue, in that order. Each digit is identified with a person. There are two types of permutations: those where Jim is in the left queue, and those where he's in the left queue. You can show a bijection between the two by taking the reversals of the permutations.
For example, if Jim is $1$, then $12345678$ is a permutation where he's in the left queue. Then $87654321$ is one where he's in the right queue. Since this is an involution, then it is a bijection, so there are as many permutations with Jim in the left queue as permutations with him in the right queue. Then the number with him in the left is half the number of all permutations, or $\frac{8!}{2}=20160$.