<Question>
Mikka wants to order a pizza with two different toppings. He has 8 different toppings to choose from. How many different pizzas could he order?
</Question>

<think>
<Plan_1>
Okay, let's see. Mikka wants to order a pizza with two different toppings, and there are 8 different toppings available. The question is asking how many different pizzas he could order. Hmm, so I need to figure out the number of possible combinations here.

First, I remember that when dealing with combinations, the order doesn't matter. Like, if he picks toppings A and B, that's the same as B and A. So it's not a permutation problem where order matters. So combinations are the way to go here.

The formula for combinations is n choose k, which is n!

<prompt_1.1> Recall the formula for combinations and explain why it is used for this problem. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
divided by (k! (n - k)! )). In this case, n is 8 because there are 8 toppings, and k is 2 because he wants two different toppings. So plugging those numbers into the formula, it would be 8! divided by (2! * (8 - 2)! ).
</execution_1.1>
</Execution_1>

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

Wait, 8! is 8 factorial, which is 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1. But since we have (8 - 2)! which is 6! in the denominator, maybe we can simplify that.

<prompt_2.1> Simplify the combination formula for 8 choose 2 and calculate the result. </prompt_2.1>

Wait, let me check if that makes sense. Let's think of a smaller example. Suppose there are 3 toppings, and you want to choose 2.

<prompt_2.2> Verify the combination formula with a smaller example of 3 toppings choosing 2. </prompt_2.2>

Alternatively, another way to think about it is: for each topping, how many choices do you have?

<prompt_2.3> Explain an alternative method to calculate the number of combinations for 8 toppings choosing 2. </prompt_2.3>
</Plan_2>

<Execution_2>
<execution_2.1>
So 8! divided by 6! would be 8 × 7 × 6! divided by 6! , which cancels out the 6! . So that leaves us with 8 × 7 divided by 2! . And 2! is 2 × 1, which is 2. So 8 × 7 is 56, divided by 2 is 28. So 28 combinations.
</execution_2.1>

<execution_2.2>
Then the combinations should be 3. Using the formula, 3! /(2! 1! ) = 6/(2×1) = 3. That's correct. So applying the same logic to 8 toppings, choosing 2, it should be 8×7 divided by 2×1.
</execution_2.2>

<execution_2.3>
For the first topping, you have 8 options. For the second topping, since it has to be different, you have 7 options. So 8 × 7 = 56. But this counts each combination twice because selecting topping A then B is the same as B then A. So to correct for that, you divide by 2, giving 56 / 2 = 28.
</execution_2.3>
</Execution_2>

<Plan_3>
Based on execution_2.1:
So the answer should be 28 different pizzas.

Let me verify once more with another approach.

<prompt_3.1> Verify the result using different approaches and confirm the number of combinations. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
If there are 8 toppings, the number of ways to choose 2 is calculated as combinations. Another way to remember is that the number of edges in a complete graph with 8 nodes is 8×7/2 = 28. Each edge represents a unique pair of toppings, which is a pizza. So that's another way to see it. 

Alternatively, if order mattered, it would be 8×7=56, which are the permutations. But since the pizza toppings are unordered, we need combinations, so 56 divided by 2! = 28. Yep, that's consistent. 

I don't think I missed anything here. The problem doesn't specify any restrictions, like some toppings can't be together or something. So all possible pairs are allowed. So 8 choose 2 is indeed 28. Therefore, Mikka can order 28 different pizzas with two different toppings.
</execution_3.1>
</Execution_3>

<Final_answer>
Therefore, Mikka could order $\boxed{28}$ different pizzas with two different toppings.
</Final_answer>
</think>