Sandwich decision

Probability Level 1

At a local deli, the following options are given for a sandwich:

  • Bread types: White, Rye, Wheat
  • Cheese Types: Swiss, Cheddar, Havarti, Provolone
  • Meat Types: Roast Beef, Turkey, Ham, Corned Beef, Pulled Pork

If a customer chooses exactly one of each type of item, then how many possible sandwiches can be made?


The answer is 60.

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Andy Hayes by Andy Hayes , 38, USA

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3 solutions

Andy Hayes
Dec 12, 2016

There are 3 types of bread, 4 types of cheese, and 5 types of meat. Using the rule of product , the number of possible sandwiches is:

3 × 4 × 5 = 60 . 3 \times 4 \times 5 = \boxed{60}.

41 Helpful 8 Interesting 2 Brilliant 4 Confused
Elias Myserlis
Jul 15, 2019

I’ve thought of an intuitional solution to THIS problem. We can arrange our ingredients on a 3-dimensional table and workout the combinations. By doing that, the rule of product becomes apparent for the case of combining objects from three discrete sets into triplets. 

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\documentclass[margin=5mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{matrix,calc}

\begin{document}
\begin{tikzpicture}[every node/.style={anchor=north east,fill=white,minimum width=1.4cm,minimum height=7mm}]
\matrix (mA) [draw,matrix of math nodes]
{
(Swiss,Roast Beef,White) & (Swiss,Turkey,White) & (Swiss,Ham,White) & (Swiss,Corned Beef,White) & (Swiss,Pulled Pork,White) \\
(Cheddar,Roast Beef,White) & (Cheddar,Turkey,White) & (Cheddar,Ham,White) & (Cheddar,Corned Beef,White) & (Cheddar,Pulled Pork,White) \\
(Havarti,Roast Beef,White) & (Havarti,Turkey,White) & (Havarti,Ham,White) & (Havarti,Corned Beef,White) & (Havarti,Pulled Pork,White) \\
(Provolone,Roast Beef,White) & (Provolone,Turkey,White) & (Provolone,Ham,White) & (Provolone,Corned Beef,White) & (Provolone,Pulled Pork,White) \\
};

\matrix (mB) [draw,matrix of math nodes] at ($(mA.south west)+(1.5,0.7)$)
{
(Swiss,Roast Beef,Rye) & (Swiss,Turkey,Rye) & (Swiss,Ham,Rye) & (Swiss,Corned Beef,Rye) & (Swiss,Pulled Pork,Rye) \\
(Cheddar,Roast Beef,Rye) & (Cheddar,Turkey,Rye) & (Cheddar,Ham,Rye) & (Cheddar,Corned Beef,Rye) & (Cheddar,Pulled Pork,Rye) \\
(Havarti,Roast Beef,Rye) & (Havarti,Turkey,Rye) & (Havarti,Ham,Rye) & (Havarti,Corned Beef,Rye) & (Havarti,Pulled Pork,Rye)\\
(Provolone,Roast Beef,Rye) & (Provolone,Turkey,Rye) & (Provolone,Ham,Rye) & (Provolone,Corned Beef,Rye) & (Provolone,Pulled Pork,Rye) \\
};

\matrix (mC) [draw,matrix of math nodes] at ($(mB.south west)+(1.5,0.7)$)
{
(Swiss,Roast Beef,Wheat) & (Swiss,Turkey,Wheat) & (Swiss,Ham,Wheat) & (Swiss,Corned Beef,Wheat) & (Swiss,Pulled Pork,Wheat) \\
(Cheddar,Roast Beef,Wheat) & (Cheddar,Turkey,Wheat) & (Cheddar,Ham,Wheat) & (Cheddar,Corned Beef,Wheat) & (Cheddar,Pulled Pork,Wheat) \\
(Havarti,Roast Beef,Wheat) & (Havarti,Turkey,Wheat) & (Havarti,Ham,Wheat) & (Havarti,Corned Beef,Wheat) & (Havarti,Pulled Pork,Wheat) \\
(Provolone,Roast Beef,Wheat) & (Provolone,Turkey,Wheat) & (Provolone,Ham,Wheat) & (Provolone,Corned Beef,Wheat) & (Provolone,Pulled Pork,Wheat) \\
};

\draw[dashed](mA.north east)--(mC.north east);
\draw[dashed](mA.north west)--(mC.north west);
\draw[dashed](mA.south east)--(mC.south east);
\end{tikzpicture}


\end{document}

(A big thanks to AJN, for the help provided for the documentation on the StackExchange forum (Link: https://tex.stackexchange.com/questions/300109/simple-visualization-of-3d-matrix).)

0 Helpful 8 Interesting 2 Brilliant 5 Confused
Ardell Deliz
May 29, 2021

Each type of bread has 20 possible ways of building a sandwich when only one ingredient from each list is allowed. With only three types of bread, the answer is 3 x 20 = 60 ... or you can multiply each list with each other 3 x 4 x 5 = 60

1 Helpful 0 Interesting 0 Brilliant 0 Confused

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