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!
 A: 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 $7\cdot6\cdot5-5\cdot4\cdot3=210-60=150$.
A: This problem is a direct application of the Subtraction Principle. We first count the number of possible ways to select a chairperson, secretary, and treasurer. Since in total there are $7$ people we can select a chairperson in $7$ ways. Once this has been done we can select a secretary in $6$ ways. Once this has been done we can select a treasurer in $5$ ways. By the Multiplication Principle we have $7\cdot6\cdot5=210$ ways to select these three positions with $7$ people under consideration, or $P(7,3)$. Now, we want to count the number of ways in which Dolph and Egbert are not allowed to be a chairperson, secretary, or treasurer. This leaves only $5$ people to work with. We can select a chairperson in $5$ ways. Once this has been done we can select a secretary in $4$ ways. Once this has been done we can select a treasurer in $3$ ways. By the Multiplication Principle we have $5\cdot4\cdot3=60$ ways to select these three positions when Dolph and Egbert aren't allowed to be selected, or $P(5,3)$. Thus by the Subtraction Principle we have $P(7,3)-P(5,3)=210-60=150$ ways for Dolph and Egbert to hold at least one of the positions.
