There are $26$ students in a homeroom. How many different ways can they be chosen to be elected President, Vice President, Treasurer, and Secretary? $\newline$ Answer:

Full solution

Q. There are

26

students in a homeroom. How many different ways can they be chosen to be elected President, Vice President, Treasurer, and Secretary?

\newline

Answer:

Understand the problem: Understand the problem. $\newline$ We need to find the number of different ways to choose $4$ students from a group of $26$ for the positions of President, Vice President, Treasurer, and Secretary. Since the order in which we choose the students matters (because the positions are distinct), we will use permutations to solve this problem.
Calculate the permutations: Calculate the number of permutations. $\newline$ The number of ways to choose $4$ students out of $26$ for distinct positions is given by the permutation formula, which is $P(n, k) = \frac{n!}{(n - k)!}$ , where $n$ is the total number of items to choose from, $k$ is the number of items to choose, $n!$ denotes the factorial of $n$ , and $(n - k)!$ is the factorial of $(n - k)$ . $\newline$ In this case, $n = 26$ and $26$ $0$ .
Perform the calculation: Perform the calculation. $\newline$ $P(26, 4) = \frac{26!}{(26 - 4)!}$ $\newline$ $= \frac{26!}{22!}$ $\newline$ $= \frac{26 \times 25 \times 24 \times 23 \times 22!}{22!}$ (since the factorials cancel out) $\newline$ $= 26 \times 25 \times 24 \times 23$ $\newline$ $= 358800$