Combinatorics Redux Followup: 2 Same, 1 Different Toppings on 3 Pizzas

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The discussion explores the combinatorial problem of selecting toppings for three pizzas, specifically focusing on the scenario where two pizzas have the same topping and one has a different topping. It establishes that there are 2^9 ways to choose toppings for a single pizza and calculates the combinations for three pizzas with different toppings using the formula 2^9C3. The main query revolves around finding the correct formula for the case of two same and one different topping, proposing that it should be 2^9*(2^9-1). However, there is confusion regarding the inclusion of a factorial in the final answer, leading to a discussion about whether the calculation is missing a factor of 2!. The thread concludes with a request for clarification on the correct approach to this combinatorial scenario.
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Followup:

Suppose we have 2^9 ways of choosing topics on a pizza. Then, if we want different topics on 3 pizzas, we can do that in 2^9C3 = (2^9*(2^9)-1*(2^9)-2)/3!

and if we want the same toppings on all the 3 pizzas, we already know that we can do that in 2^9 ways.

But what about the third option:

we want 2 same and one different topping for the 3 pizzas.

What formula will we use?

The answer is supposed to be 2^9*(2^9)-1
 
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The situation is equivalent to having to choose two different toppings; the two same toppings can be considered to be one toping for one pizza.
 
Right, in that case, we should have 2^9C2 ways of doing this.. which is (2^9*(2^9)-1)/2! but the answer doesn't mention the 2!

Am I missing something??
 
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