Welcome back, cryptocampers & co.!
Today we learned how to count! Yes, you read that right. We started off the day with a fantastic lesson on the good ol’ P(n,k)=(n!)/((n-k)!) with the one and only Dr. Woodard. Doesn’t sound like what you learned on Sesame Street? Allow us to explain. Rather than traditional 1, 2, 3, we learned to count the total number of possible combinations of distinct objects in a known set. This is part of the branch of mathematics called combinatorics.
First, we worked with permutations. Permutations count ordered arrangements of a certain number of objects from a larger pool of objects. Imagine you harvested 3 eggs (one blue, one green, and one purple) from your chickens this morning, and you have 5 friends (Allie, Bobby, Charlie, Danny, and Esther) eagerly awaiting them. They aren’t picky--they’ll take any color egg they can get. If only 3 friends can get an egg, which 3 friends will you pick? Will Allie get the green egg, Charlie get the blue egg, and Esther get the purple egg? Will Bobby get the blue egg, Danny get the green egg, and Allie get the purple egg? How many total combinations of friends and eggs exist? This is where permutations come in. Picture 3 slots--one for each egg--with a different friend filling each. Let’s start with the blue egg. How many friends are eligible to receive this egg? All 5! So there are 5 possible options for friends who can receive the blue egg. How many options are we left with for the green egg? If one friend has already received an egg, we are left with 4. Follow the same logic for the purple egg, and we are left with 3 options (see below for a visual). To calculate a permutation, we multiply these options together. Using formal notation, P(5,3)=5(4)(3)=60. We have 60 combinations of friends and eggs! How will you ever choose?
Next, we moved on to combinations. Combinations are similar to permutations, except they do not take the order of the selected objects into account, only what objects are chosen. In this eggy scenario, we count combinations if we only care about which friends receive eggs, not the order in which they receive them (i.e. we are disregarding the colors of the eggs). To calculate a combination, you first find the permutation of a certain number of ordered objects. You then divide that number by the factorial of the number of “slots” (in this case, 3). Denoted n!, a factorial means multiplying all the positive integers from 1 to n. For instance, 3!=3(2)(1)=6. This removes all redundant sets since order does not matter for combinations. For example, {Allie, Bobby, Charlie} is the same set as {Bobby, Charlie, Allie} even though they are in different orders. A combination accounts for this equivalency and removes the duplicate sets. So, what is the combination of our earlier friends and eggs scenario? Let’s calculate it. C(5,3)=(P(5,3))/3!=60/6=10.
After a brief nap and a break for lunch, Dr. Millichap sent us on quite the adventure with a cryptology-themed scavenger hunt around campus. We were split into teams of 3 and sent on our way to solve the mystery of where Dr. Woodard’s chickens went (Disclaimer: they were not his actual chickens, unfortunately.🙁).
With 6 different challenges, we practiced all the skills and ciphers we have learned throughout the week on campus as one last finale before our venture to London begins!! To recap, we went through Monoalphabetic Substitution Ciphers, Columnar Transpositions, One Time Pads, Caesar Shifts, and the Playfair Cipher. We even got a little treat as Dr. Millichap created his own code for us to solve based on our class textbook titled “The Code Book.” All groups successfully broke the codes and discovered the location of the chickens, which were held captive in Dr. Woodard’s office by a ferocious army of rubber ducks. What fowl play!
Tonight, we are all prepping for our travel day tomorrow. Various groups went out for our last Greenville dinners of this MayX, did some last minute laundry, and played the game of “will-my-suitcase-close-if-I-sit-on-it?” A classic.
Until tomorrow, cryptocampers! We’ll see you in the airport!
Ava Kinghorn & Kali Javetski
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