How many ways for 2 people not standing beside each other, + Angela must be in front of Tony? > Tony, Angela, and four of their friends are at an amusement park. They all get in line for a roller coaster. If the six of them line up at random, how many ways that Tony and Angela are NOT standing beside each other AND Angela is always in front of Tony in the line? I thought as follows: We got $6!$ ways to place the friends randomly. There are $2 \cdot 5!$ ways to put Angela and Tony together. We thus get $6! - 2 \cdot 5!= 4 \cdot 5!$ ways to answer the first question. As we have to put Angela before Tony in the line, this leaves $\frac{4 \cdot 5!}{2} = 2 \cdot 5!$ ways. Is this correct?

Yes, your answer of $$\frac{1}{2}(6! - 5!2!)$$ is correct. I would have approached the problem in the same way.

We can confirm your answer with a different approach. Observe that since there are a total of six people and Angela must appear before Tony without being adjacent to Tony, she must be in one of the first four positions.

* If she is in the first position, he must be in one of the last four positions.
* If she is in the second position, he must must be in one of the last three positions.
* If she in the third position, he must be in one of the last two positions.
* If she is in the fourth position, he must be in the last position.



Thus, there are $4 + 3 + 2 + 1 = 10$ ways to place Angela and Tony. For each way they can be arranged, the remaining people can be arranged in $4!$ ways. Hence, the number of admissible arrangements is $$(4 + 3 + 2 + 1)4! = \frac{1}{2}(6! - 5!2!)$$