<Question>
In how many ways can  5 students be selected from a group of 6 students?
</Question>

<think>
<Plan_1>
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 here. Wait, combinations because the order in which we select the students doesn't matter, right? Like, if we're just picking a group, not assigning positions or anything.

So, the formula for combinations is n choose k, which is n!

<prompt_1.1> Recall and write down the formula for combinations using factorials. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
divided by (k! times (n - k)! )). Let me make sure I remember that correctly. So in this case, n is 6 because there are 6 students, and k is 5 because we want to select 5 of them.

Plugging those numbers into the formula, it would be 6! divided by (5! times (6 - 5)! ).
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
Let me compute that step by step.

<prompt_2.1> Calculate the number of combinations using the combination formula for n=6 and k=5. </prompt_2.1>

Wait, but let me think again. Another way to check is that choosing 5 out of 6 is the same as leaving out 1 student. So instead of calculating 6C5, maybe I can calculate 6C1, which is the number of ways to choose 1 student to leave out. Since there are 6 students, there are 6 ways to choose which one to leave out.
Therefore, both methods should give the same result, and they do. That makes sense because each way of leaving out one student corresponds exactly to selecting the other five. So whether I compute 6C5 or 6C1, I get 6.

Let me verify with the combination formula again.

<prompt_2.2> Re-calculate using the combination formula for n=6 and k=1 to verify the result. </prompt_2.2>
</Plan_2>

<Execution_2>
<execution_2.1>
First, 6! is 6 factorial, which is 6 × 5 × 4 × 3 × 2 × 1. But when I divide by 5! , which is 5 × 4 × 3 × 2 × 1, those terms will cancel out. Let me write that out: 6! / (5! * 1! ) = (6 × 5 × 4 × 3 × 2 × 1) / (5 × 4 × 3 × 2 × 1 × 1). So all the terms from 5 down to 1 in the numerator and denominator cancel out, leaving just 6/1, which is 6.
</execution_2.1>

<execution_2.2>
6! is 720. 5! is 120, and (6 - 5)! is 1! which is 1. So 720 divided by (120 times 1) is 720 / 120 = 6.
</execution_2.2>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2:
I don't think there's any mistake here. The key was recognizing that choosing 5 out of 6 is equivalent to choosing 1 out of 6 to leave out, which simplifies the calculation.
So the answer should be 6 ways.

Wait, just to make sure there isn't a trick here. The problem says "selected from a group of 6 students". Are there any restrictions?

<prompt_3.1> Verify if there are any restrictions in the problem that could affect the calculation. </prompt_3.1>

Another check: if I list all possible groups of 5 from 6 students. Let's say the students are A, B, C, D, E, F. The possible groups are:

<prompt_3.2> List all possible groups of 5 students from a group of 6 to verify the result. </prompt_3.2>
</Plan_3>

<Execution_3>
<execution_3.1>
Like, do we need to consider the order? But no, because combinations are order-independent. So 6C5 is correct. Alternatively, if it were permutations, the number would be higher, but since it's selection without order, combinations are the way to go.

Alternatively, if I use the combination formula directly: 6C5 = 6! / (5! * (6-5)! ) = 6. So yes, that's correct.
</execution_3.1>

<execution_3.2>
1. A, B, C, D, E
2. A, B, C, D, F
3. A, B, C, E, F
4. A, B, D, E, F
5. A, C, D, E, F
6. B, C, D, E, F

That's 6 groups, each missing one of the six students. So yes, that's 6 different ways.
</execution_3.2>
</Execution_3>

<Final_answer>
Therefore, the number of ways to select 5 students from a group of 6 is $\boxed{6}$.
</Final_answer>
</think>