1. **Problem statement:** We need to select a group of 5 people, but 3 places are already taken. We want to find in how many ways the other 2 places can be filled using permutations and combinations. 2. **Understanding the problem:** Since 3 places are already taken, we only need to select 2 more people to complete the group of 5. 3. **Using combinations:** If the order of selection does not matter, we use combinations. The number of ways to choose 2 people from the remaining candidates is given by the combination formula: $$ C(n, r) = \frac{n!}{r!(n-r)!} $$ where $n$ is the total number of candidates available to fill the remaining 2 places, and $r=2$. 4. **Using permutations:** If the order of selection matters (i.e., who is chosen first and second), we use permutations. The number of ways to arrange 2 people out of $n$ candidates is given by the permutation formula: $$ P(n, r) = \frac{n!}{(n-r)!} $$ where $r=2$. 5. **Final step:** Since the problem does not specify the total number of candidates available to fill the remaining 2 places, the formulas above show how to calculate the number of ways depending on whether order matters or not. **Summary:** - Number of ways to select 2 people without order: $$ C(n, 2) = \frac{n!}{2!(n-2)!} $$ - Number of ways to select 2 people with order: $$ P(n, 2) = \frac{n!}{(n-2)!} $$ Replace $n$ with the actual number of candidates available to get the numerical answer.