Are there 3000 types of sandwiches?

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The discussion revolves around a delicatessen's claim of offering 3000 types of sandwiches based on various combinations of meats, cheeses, breads, and condiments. One participant calculated a total of 3072 combinations, initially supporting the deli's claim, but later reconsidered the requirement of having at least one meat or cheese, leading to a revised total of 2032 combinations. This adjustment indicated that the deli's advertisement is misleading. The final consensus suggests that the deli's claim is false, as the actual number of sandwich combinations does not meet the advertised figure. The discussion highlights the importance of considering all conditions when calculating possible combinations.
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Homework Statement


A delicatessen advertises that if offers 3000 types of sandwiches. If at this deli it is possible to have any combination of salami, turkey, bologna, corned beef, and ham with or without Swiss and/or American cheese on French, white or whole wheat bread, and possible additions of lettuce, tomato, and mayo, is the deli's advertisement true? Assume that a sandwich necessarily has bread and at least one type of meat or cheese.


Homework Equations





The Attempt at a Solution


I got 3072 as my solution, so the deli is being truthful.
My reasoning was that for the main topping there are 2^7 choices. I then multiplied that by 3, because there are 3 types of breads, but you can only have one of them. Then 2^3 different types of condiments.

However, knowing that the sandwich has to have at least one meat or cheese, that would make the choices down to 2^6 I think. Not sure :/
 
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Hi Intervenient! :smile:

Intervenient said:

Homework Statement


A delicatessen advertises that if offers 3000 types of sandwiches. If at this deli it is possible to have any combination of salami, turkey, bologna, corned beef, and ham with or without Swiss and/or American cheese on French, white or whole wheat bread, and possible additions of lettuce, tomato, and mayo, is the deli's advertisement true? Assume that a sandwich necessarily has bread and at least one type of meat or cheese.

Homework Equations


The Attempt at a Solution


I got 3072 as my solution, so the deli is being truthful.
My reasoning was that for the main topping there are 2^7 choices. I then multiplied that by 3, because there are 3 types of breads, but you can only have one of them. Then 2^3 different types of condiments.

However, knowing that the sandwich has to have at least one meat or cheese, that would make the choices down to 2^6 I think. Not sure :/
If I understand your problem statement correctly, you need at least 1 choice from the first 7 main toppings.
Or put otherwise, it is not allowed to have none of the first 7 main toppings.

Si there are 2^7 combinations of which exactly 1 is not allowed.
 
okay so I got a different #
(2^7 -1) for meat / cheese selections (-1 it says at least one meat of cheese)
2 for bread choice
2^3 for condiments
---------------------
2032 choices

so that answer is no - "pants on fire"

** pants on fire: referring to how PolitFact rates politician statements true , partially true, false and pants on fire for really really false)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks

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