Okay, let's try to figure out how many different pizzas Mikka can order. He wants a pizza with two different toppings, and there are 8 toppings available. Hmm, so first, I need to remember how combinations work because the order of toppings doesn't matter here. Like, a pizza with pepperoni and mushrooms is the same as a pizza with mushrooms and pepperoni, right? So it's not a permutation problem where order matters. 

Wait, so combinations are used when the order doesn't matter. The formula for combinations is C(n, k) = n! / (k! * (n - k)!), where n is the total number of items, and k is the number of items we're choosing. In this case, n is 8 toppings, and k is 2 toppings. So plugging in the numbers, that would be C(8, 2) = 8! / (2! * (8 - 2)!).

Let me calculate that step by step. The factorial of 8 is 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1, which is 40320. But the denominator is 2! times 6!, which is 2 × 720. Wait, 6! is 720? Let me check that. 6×5×4×3×2×1 is 720, yes. So 2! is 2, so 2 × 720 = 1440. Then 40320 divided by 1440. Let me do that division. 40320 ÷ 1440. Hmm, 1440 × 28 is 40320, because 1440 × 10 is 14400, ×20 is 28800, then ×8 is 11520, adding those gives 14400 + 11520 = 25920, which is 28 × 900. Wait, maybe I should divide 40320 by 1440. Let's see, 1440 × 28 = 40320. So the answer is 28. 

Wait, does that make sense? Let me think again. If there are 8 toppings, how many ways can you pick 2? Let's list them out. For 8 toppings labeled A to H, the combinations would be AB, AC, AD, AE, AF, AG, AH, BC, BD, BE, BF, BG, BH, CD, CE, CF, CG, CH, DE, DF, DG, DH, EF, EG, EH, FG, FH, GH. Let's count them. 

Starting with A: 7 combinations (AB, AC, AD, AE, AF, AG, AH). Then B: 6 (BC, BD, BE, BF, BG, BH). Then C: 5 (CD, CE, CF, CG, CH). D: 4 (DE, DF, DG, DH). E: 3 (EF, EG, EH). F: 2 (FG, FH). G: 1 (GH). H: 0. So total is 7+6+5+4+3+2+1 = 28. Yep, that matches the combination formula. So the answer is 28 different pizzas.

I think that's correct. Another way to think about it is that for the first topping, you have 8 choices, and for the second topping, you have 7 choices. That gives 8×7=56. But since the order doesn't matter, each pair is counted twice (once as AB and once as BA). So you divide by 2 to eliminate the duplicates. 56/2=28. That's another way to get the same answer. So both methods confirm that the number of different pizzas Mikka can order is 28.
Therefore, Mikka can order $\boxed{28}$ different pizzas.
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