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

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The discussion focuses on combinatorial calculations for selecting toppings on three pizzas, specifically addressing the scenario of having two identical toppings and one different topping. The formula for this configuration is derived as 2^9 * (2^9 - 1), where 2^9 represents the total number of topping combinations. The participants clarify that the two identical toppings can be treated as one choice, leading to the calculation of combinations using 2^9C2, which simplifies to (2^9 * (2^9 - 1)) / 2!. However, the discussion notes a discrepancy regarding the inclusion of the 2! factor in the final answer.

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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??
 
Last edited:

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