## Wednesday, April 13, 2011

### Chipotle Math

A friend who is living abroad told us that what he misses most about the US is Chipotle Mexican food. Curiously, we checked out the new branch that recently opened in our neighborhood and quickly became hooked. It seems very simple: tortilla or taco and a few basic but delicious ingredients:

But the combination of flavors, the freshness of ingredients, the spectrum of spiciness that you can choose from, and just the right proportions of everything make it a special comfort meal.

We noticed that there is claim on some of Chipotle paper cups and wrapping paper that "there are 60-some-thousand flavor combinations." 60,000 is quite a large number. Do you think it is really possible with all these ingredients pictured above?

Answers accepted all day long on Friday April 15th and Saturday April 16th, on our Family Puzzle Marathon. They will be hidden until Sunday morning (EST) and everyone who solved it will get a puzzle point. Please, explain your answer.

Kim said...

There are 16 ingredients (assuming we include both black and pinto beans, and all four salsas), and a customer can include as many or as few as these as they would like. For the sake of simplicity, I will assume they can choose any ingredient independently of all other choices (that is, they don't have to choose between chicken and steak, they can have both and they can have multiple salsas).

In that case, you would calculate it as 2^16 because each choice is either "on" or "off", which is 65536. You might argue that the choice of having none of these is not plausible (as is probably the choice of just having sour cream), but let's ignore these for the sake of simplicity. Also, we could complicate the calculation by making people choose one meat or one salsa, but that's no fun!

Of course, we could further complicate things by allowing that there is also a choice of whether or not to put these in a tortilla, so the tortilla itself becomes another ingredient. And maybe there is more than one kind of tortilla? Maybe all choices come in small or large?

In any event, I think they can defend the "there are 60-some-thousand flavor combinations" statement.

Unknown said...

The simple answer is yes, with a few assumptions. If we assume that there are 12 ingredients, because of 12 pictures, then no. But at closer inspection of the pictures there are 2 kinds of beans and 4 salsas - so a total of 16 ingredients. Assuming that a flavor combination can have anywhere from 1 to 16 ingredients, we would use the following formula: n!/(k!*(n-k)!)
n is 16 in all cases. We then need to calculate this for all values of k from 1 to 16.

k= 1 = 16
k= 2 = 120
k= 3 = 560
k= 4 = 1,820
k= 5 = 4,368
k= 6 = 8,008
k= 7 = 11,440
k= 8 = 12,870
k= 9 = 11,440
k= 10 = 8,008
k= 11 = 4,368
k= 12 = 1,820
k= 13 = 560
k= 14 = 120
k= 15 = 16
k= 16 = 1

Then we add up all the values and get a total of 65,398 flavor combinations. Although one would wonder if practically they would serve just lettuce and salsa...or just a combination of all 4 salsa. So, even though there are that many flavor combinations they would not necessarily serve that many.

PApaige said...

No, we do not think 60,000 combinations are possible. We applied the "handshake" theory to all the fillings and doubled for the two shells - taco vs. tortilla - and were not even close to 60,000. Then we tried the different combinations allowing for more than 2 toppings and/or fillings and at least got into the thousands, but still not close to 60,000.

Wang said...

There are 2 bases and 12 different ingredients.

Let's start with the different ingredients first - how many different combinations are there?

You can first, have no toppings. 12C0 (there's 1 way to have this)

Next, you can have one topping. 12C1 (there's 12 ways to have this)

Next, you can have two toppings. 12C2 (there's 66 ways to do this). I use 12C2 because getting beans and guacamole is the same as getting guacamole and beans.

You can then see that the number of different combinations is 12C0+12C1+12C2+...+12C12

in other words, 2 ^ 12 = 4096.

Even if we say that the two bases (tortilla or taco) make a different taste, we only have 8192 different flavors in total.

If they think that different portions of everything would make different flavors then yes, there would be more than 60,000 flavors (a small dollop of guacamole certainly would taste differently than a large dollop). Perhaps they were thinking this way rather than purely on ingredients.

John Golden said...

The maximum choices is usually a simple power of 2. For each item, on or off. I think they're a little short on 60,000. 16 items pictured, plus maybe another kind of cheese, so 17. 2^17 = 131,072. But that includes 8,192 with all four meats! If you choose one meat, 32,768, for 1C4*2^13. 16K if only one kind of cheese. But it'd be cool if they sold a maximum burrito with yes to everything!

anne-marie said...

There are 12! possiblilities of having a unique sandwish .
479001600 flavor combinations =12*11*10*9*8*7*6*5*4*3*2

Bean said...

Oh, how I wish I remembered my High School math! I know there is a beautiful formula out there for this, but here I am hacking at it with drawings on paper.

Yes, there are 65,504 combinations. You need to count each salsa separately, and each kind of bean separately, for a total of 16 ingredients.

So, having forgotten everything I ever learned about combinations, I had to make up the formula from scratch. I reason that there is only one way to get a plain tortilla; there are 16 ways to get a tortilla with a single ingredient; 16x15/2 ways to get a two-ingredient burrito. To combine three things you use 16x15x14, but then you have to divide by 3x2 to get rid of all the duplicate combinations. This pattern continues until you are dividing the 16! combinations by 16!, which makes sense...there is only one way to get an "everything" burrito. So the eight ingredient burrito will have the most variations, but the burritos with very few or very many ingredients should have fewer alternatives.

So my formula for each term turns out to be x/(((x-n)!)x!), where n is the number of the term and x is the number of items. You need to add those all up to get your answer. I crunched it out for 16, and got 65,504. I suppose you could also count with/without tortilla as an option, and work with 17 as your number. But 60,000 is fair enough.

Pat said...

Yes, 60,000 is reasonable; I count 65,636.

There are 4 choices of salsa shown, 2 choices of beans, and 10 other items. This mean there are 16 things that you can either have or not have on your burrito. For each of the 16 items there are 2 choices, meaning 2^16th combinations.

---
I'm a professor dealing with manufacturing issues in Detroit,and we discuss this when thinking about auto production. From body styles to engines to interior/exterior colors, to seats, to radios, etc. there are probably 50 different option dimensions, each with multiple choices. One of the makers advertised that there were more possible options than there are drivers in the US.

At Chipotle, all the options are in front of the person building your burrito, and s/he just adds things as you ask for them. In auto assembly that's not possible, so every optional item is sequenced so that the operator can always take the top thing off the stack. If there are 50 different places where things need to be sequenced properly, and something gets out of sequence, the problem will just continue until generally some sensor figures it out. You can have situations where the left side of the vehicle says it's an F250, while the right says it's an F350.

TyYann said...

Well, each Chipotle can be made from 0 (or 1, but that doesn't make a lot of difference) to 12 ingredients.
If I choose 3 ingredients, I have the combination of 3 among 12 ways to choose, witch is 220.
Here's the list:
0 ingredient: 1 possibility
1 ingredient: 12 possibilities
2 ingredients: 66 possibilities
3 ingredients: 220 possibilities
4 ingredients: 495 possibilities
5 ingredients: 792 possibilities
6 ingredients: 924 possibilities
7 ingredients: 792 possibilities
8 ingredients: 495 possibilities
9 ingredients: 220 possibilities
10 ingredients: 66 possibilities
11 ingredients: 12 possibilities
12 ingredients: 1 possibility.
The sum being 4096 (or 4095 if 0 ingredient is not considered valid)
So, this doesn't look possible except if you think that flavour is not only the ingredients, but also the amount of them compare to the others...

Anonymous said...

That 60,000 number of combinations surprised me so I tried to find out how many different combinations there could be. When I first looked at the menu I saw there were 12 menu items. I then use the Combination formula (ex. 12!/9!). I ended up with 4095 possible combinations.
I went back to the menu and realized there were 4 different salsas so I tried it again, using 15 different combinations and ended up with 32,763 different combinations.
The breakdown looks like this:
any 1 item - 15 combos
any 2 items - 105 combos
any 3 items - 455 combos
any 4 items - 1363 combos
any 5 items - 3003 combos
any 6 items = 5005 combos
any 7 items - 6435 combos
any 8 items - 6435 combos
any 9 items - 5005 combos
any 10 items - 3003 combos
any 11 items - 1363 combos
any 12 items - 455 combos
any 13 items - 105 combos
any 14 items - 15 combos
any 15 items - 1 combo
So, I came up with 32,763 different combos but some of those combos are made up of one salsa and nothing else or two salsas and nothing else. Another possibility would be guacomole and sour cream and nothing else. Yum!
math mover

Annie said...

I would say that, no, there can not be 60,000 combinations from 12 food items.

I immediately thought of factorials when reading the problem but had to brush up a little. This is not a straight permutation problem so the solution is not equal to 12!. The equation I used is:

nCm= n(n-1)(n-2).....(n-m+1) / m(m-1)...1 where n = 12 food items and m=# items in a combination

For one item chosen there would be 12 combinations, for 2 items chosen there would be 66 combinations:

12C2 = 12*11/ 2*1 = 132/2=66

For 3 items =220 combinations
4 " = 495 "
5 " =792 "
6 " =924 "
7 " =792 "
8 " =495 "
9 " = 220 "
10 " =66 "
11 " =12
12 " =1

Total combinations is 4095, far less than the 60,000 advertised. Am I in the ballpark???????

Maria said...

You all deserve a burrito or tacos today, and a puzzle point.
I would agree with many of you that said that Chipotle can defend their math and surprisingly this limited selection of ingredients allows to create over 60,000 different flavours. And the simplest explanation is that if we line all 16 (!) ingredients up and like in a binary system use 1(include) or 0(exclude) we will have 16 digit long binary number. How many different numbers we can create - 2 in the power 16, equal 65,536.

Of course there are some combinations that are questionable. Just sour cream or just salsa in a tortilla, but I know some kids who will gladly eat that. Combining salsas or combining meats - I agree that we should exclude some unrealistic combinations like all 3 meats and nothing else, or 2-4 salsas and nothing else but barbacoa, chicken and salsa may be work for some.

And if we add a choice of tortilla and soft or hard taco shells, we'll need to multiply our result by 3 and easily get well above 60,000

Chipotle know their math!

Maria said...

Bean - this was your 10th puzzle point. Congratulations! Prepare to be wrapped into a puzzle next week.

DJ Emir Mixtapes And Designs said...

They have a choice of Hard tacos, Soft Shell Tacos, Flour Tortilla or Bowl... Bowl meaning no tortilla which would be considered a flavor variant. You can also add chips to further complicate things, so I'd say yes well over.

vinoo cameron said...

In choices like these you have to figure out what the rational choices are and what the non-rational choices are. Choice with groups of 3 choices each, the rational choice will be much less. As an instance if the choices were in groups 0f 3, the choices will be limited, but since 1 is a prime number the choices are very much finite, provides each tortilla will have to have minimum number of ingredients specified , which has not been done here in this math problem.

May be I am confused here, generally the formulas are correct here, but if all 16 choices had to be part of a a tortilla, there will be one choice.In this problems the choices are finite and calculable as such, above is correct