A student formed a club at their school. They have 13 members, and need to elect a president, vice president, and treasurer. How many ways are there to fill these officer positions?

Hey, everyone. In this problem, we're told that a student has formed a club at their school. They have 13 members, but they need to elect a president, a vice president, and a treasurer out of those 13 members, so we want to know how many different ways there are to fill these 3 officer positions. Now we're going to be working with permutations here out of a total group of 13 choosing only 3, so let's go ahead and get started. Now the first thing we want to do is identify what our values for n and r are before we use our permutations formula. So here remember our total is always going to be our value for n. This is going to be our highest number, so this 13 is my value for n. Then I have 3 different positions: president, vice president, and treasurer. So that tells me that my value for r is going to be 3. From here we can go ahead and set up our permutations formula with our 13 on this left side, our p in the middle to represent our permutations, and then our value for r on the other side. Now when setting up our formula, remember that your value for n always goes on the top. This is going to be your highest number, and we're always going to start with that. So here we have 13 factorial on the top and then on the bottom we do n minus r factorial, in this case 13 minuteus 3 factorial. From here we can go ahead and solve this the way that we know how. So we can't simplify the top anymore, we're just going to leave that as 13 factorial for now, But on the bottom, we have 13 minuteus 3 factorial, which is just 10 factorial. Now we know that we can rewrite this numerator in order to cancel out our denominator, so let's go ahead and do that. I can rewrite 13 factorial as 13 times 12 times 11 times 10 factorial, which is exactly what I wanna cancel out on the bottom, 10 factorial. So that 10 factorial is going to cancel, leaving me with just 13 times 12 times 11, which multiplying that out gives me 1,716. So my final answer is that there are 1,716 different ways to fill these 3 officer positions. Thanks for watching, and I'll see you in the next one.