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Topic: Ways to make a Burrito (Read 1075 times) |
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Virgilijus
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Ways to make a Burrito
« on: Aug 26th, 2015, 12:28am » |
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After winning his court case due to a procedural error, Willy Wutang has decided to celebrate by having some delicious Mexican food.
Unknown to Willy, an old and justifiably vengeful ex is one of the lead chef's in the kitchen and she has plans for Willy's food; instead of layering all the ingredients in the middle of the burrito, she will place all the ingredients in a line. Devious and, most importantly, legal.
However, there's a stipulation; of the six ingredients, there are some that, by tradition, should never touch each other:
Avocados and Beans
Cheese and Diced Peppers
Extra Spicy Hot Sauce and Fried Rice
If this is the case, how many different types of burritos could the vengeful ex make for Willy? (Note: philosophically, burrito ACBEDF is considered the same as burrito FDEBCA)
Spurned Ex
https://i.imgur.com/ILfmnSZ.jpg
Example of a good burrito (from left to right, AEBDFC)
https://i.imgur.com/dwjsfar.jpg
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| « Last Edit: Aug 26th, 2015, 12:29am by Virgilijus » |
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rmsgrey
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Re: Ways to make a Burrito
« Reply #1 on: Aug 26th, 2015, 6:17am » |
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Reasoning:
| hidden: | If you just build the burrito from left to right, there are two possible patterns for the first three ingredients - either the third ingredient is the first's partner, or it's one of the remaining pair.
In the former case, the number of possible burritos is 6*4*1*2*1*1; in the latter case, it's 6*4*2*2*2*1.
Since you can reverse any burrito, and get one equally good, the final total is half that, or 120 |
Answer: 120
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Virgilijus
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Re: Ways to make a Burrito
« Reply #2 on: Aug 26th, 2015, 9:30am » |
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Correct!
I knew it would be solved quickly on here. When my friend and I solved it, we had to do it a little more visually and thought of it like this: if you put pairs of ingredients on opposite faces of a die, how many ways can you travel across the entire surface of the die if you can only go through each face one time?
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rloginunix
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Re: Ways to make a Burrito
« Reply #3 on: Aug 26th, 2015, 10:05am » |
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Interesting equivalence idea.
I'd stick with rmsgrey's clean academic approach - partition of a set into pairwise disjoint subsets - to use the Addition Counting Principle and then the independence of the number of choices in a series of consecutive decisions - to use the Multiplication Counting Principle.
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