1. **Problem Statement:** My mom needs help to recover her phone with a 3-digit passcode. The only numbers she remembers are 1, 6, and 8. We need to find how many possible 3-digit passcodes can be formed using these digits. 2. **Identify values of n and r:** - $n$ is the total number of digits to choose from, which is 3 (digits 1, 6, and 8). - $r$ is the number of digits in the passcode, which is also 3. 3. **Permutation formula:** The number of permutations of $n$ items taken $r$ at a time is given by: $$P(n,r) = \frac{n!}{(n-r)!}$$ 4. **Solution:** Calculate $P(3,3)$: $$P(3,3) = \frac{3!}{(3-3)!} = \frac{3!}{0!}$$ Recall that $0! = 1$, so: $$P(3,3) = \frac{3 \times 2 \times 1}{1} = 6$$ 5. **Final answer:** There are **6** possible 3-digit passcodes that can be formed using the digits 1, 6, and 8. 6. **Explanation:** Since the passcode uses all three digits without repetition, the number of different ways to arrange these digits is the number of permutations of 3 digits taken 3 at a time, which is 6. Each arrangement corresponds to a unique passcode.