Okay, let me try to figure out how many ways we can select 5 students from a group of 6. Hmm, so the problem is asking for the number of combinations, right? Because the order in which we select the students doesn't matter here. Let me recall the formula for combinations. It's n choose k, which is n! divided by (k! times (n-k)!). 

So in this case, n is 6 because there are 6 students, and k is 5 because we need to choose 5 of them. Let me write that down: 6 choose 5. Using the formula, that would be 6! divided by (5! times (6-5)!). Simplifying the denominator, (6-5)! is 1!, which is 1. So the denominator becomes 5! times 1, which is just 5!. 

Now, 6! is 6 times 5 times 4 times 3 times 2 times 1. But since the denominator is 5!, which is 5 times 4 times 3 times 2 times 1, we can cancel out the 5! from both the numerator and the denominator. That leaves us with 6 divided by 1, which is 6. So there are 6 ways to choose 5 students from 6. 

Wait, let me double-check that. If there are 6 students and we need to leave out 1, then the number of ways to leave out 1 student is the same as the number of ways to select 5. Since there are 6 students, each one could be the one left out. So that's 6 different ways. Yeah, that makes sense. 

Alternatively, if I list all the possible combinations, each combination would exclude one student. Since there are 6 students, each exclusion corresponds to one combination. So that's 6 combinations. 

So both methods give me the same answer. That seems right. I don't think I made any mistakes here. The key was recognizing that choosing 5 out of 6 is the same as choosing which one to exclude, which is straightforward. 

Therefore, the number of ways is 6.
</think>