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Thu 24 Aug, 2006 02:25 am
Here's a tough one, I managed to get 243 (or 244), but it was wrong apparently.
Any one want to give this one a go? =p
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In your local fast food restaurant the contents of the menu are as follows:
Bread: Brown, Rye, Italian
Meats: Bacon, Ham, Salami
Cheese: Cheddar, Blue, Cream
Topping: Olives, Lettuce, Pickles
You're told that you 1 type of bread, 1 kind of meat, 2 types of cheese and 3 types of topping per a sandwhich.
The minimum required food is bread with nothing on it.
What are the number of options you have?
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//Scorpio
Bread=b Meat =m Cheese =c Topping =t
Assuming combinations
b =3, b+m=9, b+c=9, b+2c=9, b+t =9, b+m+c=27, b+m+2c=27, b+m+t=27, b+c +t=27, b+2c+t=27, b+m+c+t=81, b+m+2c+t=81
Total = 336
3 choices for bread
3+1 choices for meat
3+3+1 choices for cheese
1+3+3+1 choices for toppings
3*4*7*8=672
Quote:Bread=b Meat =m Cheese =c Topping =t
Assuming combinations
b =3, b+m=9, b+c=9, b+2c=9, b+t =9, b+m+c=27, b+m+2c=27, b+m+t=27, b+c +t=27, b+2c+t=27, b+m+c+t=81, b+m+2c+t=81
Total = 336
Hmm...I did not spot toppings could be all three. So I need to add
b+2t =9, b+3t=3, b+m+2t=27, b+m+3t=9, b+c+2t=27, b+2c+2t =27,
b+c+3t =9 b+2c+3t+9, b+m+c+2t=81, b+m+2c+2t =81 b+m+c+3t=27
b+m+2c+3t=27
Making new total = 672 In agreement with makr !
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