1) At Joe's Pizza Parlor, in addition to cheese there are 8 : PS Archive
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# 1) At Joe's Pizza Parlor, in addition to cheese there are 8

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1) At Joe's Pizza Parlor, in addition to cheese there are 8 [#permalink]

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13 Feb 2007, 02:59
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1) At Joe's Pizza Parlor, in addition to cheese there are 8 different toppings. If you can order any number of toppings, then how many different toppings are possible?
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13 Feb 2007, 07:41

8C1+8C2+8C3+8C4+8C5+8C6+8C7+8C8+1 (no toppings)
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13 Feb 2007, 10:36
In my city you can order pizza without cheese.

So, there are 9 different topping for me.

And to answer why it is not 8!.

We can have 10 different situation here.

Case 1: have all 9 topping = 9C9
Case 2: have 8 topping = 9C8
...
Case 10: have 0 topping = 9C0 (I wouldn't order this one )

Count all possible combination
= 9C0 + 9C1 + 9C2 + 9C3 + 9C4 + 9C5 + 9C6 + 9C7 + 9C8 + 9C9

= 1 + 9 + 36 + 84 + 126 + 126 + 84 + 36 + 9 + 1
= 512

And now the shortcut.

Since: nC0 + nC1 + nC2 + ... + nC(n-1) + nCn = 2^n

There are 9 toppings so the possible outcome = 2^9 = 512

----

Well, I think the question is not clear to me whether 9 or 8 is the right topping.
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13 Feb 2007, 11:09
Im sorry im new here, what do you mean by the "c" in 9C9, 8C8, etc..
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14 Feb 2007, 02:50
Tuneman wrote:
Why wouldn't the answer be 8! ?

8! is number of permutations of 8 different things taken 8 at a time. The above scenerio is a typical Combination problem (where order is not important). This implies total no. of cominations would be the summation of indivisual combination of 8 topping taken from 0 to 8 at a time, and that is nothing but :-
C1+8C2+8C3+8C4+8C5+8C6+8C7+8C8+1 (no toppings)
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14 Feb 2007, 04:30
auniyal wrote:
Tuneman wrote:
Why wouldn't the answer be 8! ?

8! is number of permutations of 8 different things taken 8 at a time. The above scenerio is a typical Combination problem (where order is not important). This implies total no. of cominations would be the summation of indivisual combination of 8 topping taken from 0 to 8 at a time, and that is nothing but :-
C1+8C2+8C3+8C4+8C5+8C6+8C7+8C8+1 (no toppings)

Perfect!
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19 Feb 2007, 09:03
8 different toppings taken 1, 2, 3... at a time using combination.
I agree with ggarr...   [#permalink] 19 Feb 2007, 09:03
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