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Combinatorics redux

  1. Jul 4, 2007 #1

    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: Jul 4, 2007
  2. jcsd
  3. Jul 4, 2007 #2
    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.
  4. Jul 4, 2007 #3
    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: Jul 4, 2007
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