Combinatorics redux

  • Thread starter sk381
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  • #1
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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
 
Last edited:

Answers and Replies

  • #2
1,425
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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.
 
  • #3
19
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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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