myhrerd
Jan 9, 2008, 08:24 PM
If you are building hamburgers and have 8 different toppings, how many different combinations of hamburgers can you make?
twinkiedooter
Jan 9, 2008, 08:46 PM
I haven't been to Burger King in years so I'm not even going to be close with my answer. I only need one hamburger with all 8 toppings on it.
N0help4u
Jan 9, 2008, 08:55 PM
hamburger, bun and 1 topping each =8
hamburger, bun and 8 toppings =8
hamburger, bun and 7 toppings x _ different ways
hamburger, bun and 6 topping x _ different ways
hamburger, bun and 5 toppings x _ different ways
hamburger, bun and 4 toppings x _ different ways
hamburger, bun and 3 toppings x_different ways
hamburger, bun ketchup, mustard,
hamburger, bun ketchup, onion
hamburger, bun ketchup, relish
hamburger, bun ketchup, saurkraut
hamburger, bun ketchup, cheese
hamburger, bun ketchup, chili
hamburger, bun ketchup, secret ingredient
hamburger, bun mustard, onion
hamburger, bun mustard, relish
hamburger, bun mustard, saurkraut
hamburger, bun mustard, cheese
hamburger, bun mustard, chili
hamburger, bun mustard, secret ingredient
hamburger, bun &...
I got lost :(...
Plus 0NE plain!
galactus
Jan 10, 2008, 01:33 PM
The problem is rather ambiguous. I assume you can use any amount of topping. You don't have to use all 8 and order doesn't matter.
There is C(8,1)=8 ways to choose one topping
C(8,2) = 28 ways to choose 2 toppings
C(8,3)=56 ways to choose 3 toppings
C(8,4)=70 ways to choose 4 toppings
C(8,5)=56 ways to choose 5 toppngs
C(8,6)=28 ways to choose 6 toppings
C(8,7)=8 ways to choose 7 toppings
C(8,8)=1 way to choose 8 toppings
Of course, 1 way for no toppings
Add them all up and we get 256 possible hambuger combos.
andreafc
Jan 11, 2008, 02:57 AM
Galactus is right. There is a general way to get to this result with a simple formula, that works with any number in just one step. However, since I 'm too lazy to post embedded math, I attach here a pdf explaining how.