Preforming Counting Permutations Problem: A seven-person committee composed of Alice, Ben, Connie, Dolph, Egbert, Francisco, and Galvin is to select a chairperson, secretary, and treasurer. How many selections are there in which at least one office is held by Dolph or Egbert? Each person may only hold at most one office. My approach: Because the total number of permutations possible are 210 b/c the first office has 7 options, second office has 6 options, third office has 5 options 7*6*5 = 210 Then I subtract the possible offices with Dolph in an office which is 30 First office has no choice as it is Dolph so it is 1, second office has 6 choices, third office 5 choices gives me 1 * 6 * 5 = 30 Multiply 30 by 2 to account for Egbert resulting in 60. 210 - 60 = 150 So there are 150 permutations possible. Is my methodology correct? Thanks for your time!

Your "$210-60=150$" is the correct computation, and I see how you got $210$, but I can't follow your explanation of how you got $60$. What you want to subtract from the $210$ total selections is the number of selections in which neither D nor E holds an office, which means all offices are held by the other $5$ people, so the subtrahend is $5\cdot4\cdot3=60$. The final answer is $7\cdot6\cdot5-5\cdot4\cdot3=210-60=150$.