February 2025
The six sets are:
Going by the labels in the statement of the problem, the six sets are \[\{A,B,E\},\{A,F,L\},\{C,G,I\},\{D,E,L\},\{E,I,J\}, \{J,K,L\}.\]
Since each property has three options, and there are four properties, there are \[3\times3\times3\times3 = 81\] cards in a SET! deck.
For the next two questions, we first make the following observation: Given any two cards \(A\) and \(B\), there is a unique third card \(C\) so that the three cards \(A\), \(B\), and \(C\) form a set. Let’s see why this is true.
For each property, if the options on \(A\) and \(B\) are the same, then the option for that property on card \(C\) must also be the same as it is for \(A\) and \(B\). If the options are not the same for \(A\) and \(B\), then \(C\) must have the third option for that property.
For example, if \(A\) and \(B\) are both shaded solid, then \(C\) must be shaded solid. On the other hand, if the shading of \(A\) is striped and the shading of \(B\) is solid, then the shading of \(C\) must be open.
With the observation above in mind, to create a set that contains the given card, we just have to choose any other card. Then the given card with our choice of second card uniquely determines a third card that forms a set.
There are \(80\) choices for a second card. However this counts every set twice! To see why, call the card we are given in the question \(A\). Suppose we choose a card \(B\), and the unique third card that forms a set is \(C\). If instead we choose \(C\) as the second card, then \(B\) is the unique card that forms a set with \(A\) and \(C\). So choosing \(B\) as the second card results in the same set as choosing \(C\) as the second card.
Therefore the answer is \(\frac{80}{2} = 40\).
Solution 1: We again rely on the observation above that once we have chosen two cards as part of a set, the third is determined.
There are 81 ways to choose the first card in our set, and 80 ways to
choose the second. However, since there are \(3! = 6\) ways to order the three cards in a
set, the number \(81\times 80\) counts
each set six times. Therefore the number of different sets in a SET!
deck is \(\frac{81\times 80}{6} =
1080\).
Solution 2: Using Question \(3\), we know each card appears in \(40\) sets. Since there are \(81\) cards in a full deck, and \(3\) cards in a set, there are \(\frac{40 \times 81}{3} = 1080\) sets in a
SET! deck.
For this question, we will need to introduce a different way of thinking about the cards. We can encode each card by a \(4\)-tuple \((q,r,s,t)\), where each entry \(q,r,s,t\) is either \(1\), \(2\), or \(3\) as in the following table:
| Entry in the \(4\)-tuple | Number | Shading | Colour | Shape |
|---|---|---|---|---|
| \(1\) | \(1\) | open | red | diamond |
| \(2\) | \(2\) | striped | green | oval |
| \(3\) | \(3\) | solid | purple | squiggle |
For example, consider the cards from one of the sets from Problem 1 above:
These cards are encoded by the tuples \((3,1,1,3)\), \((1,2,1,1)\), and \((2,3,1,2)\) respectively.
Let’s think about what it takes for three \(4\)-tuples to form a set. Each coordinate in the tuple (ie, the first, second, third, or fourth entry) corresponds to one of the properties (number, shading, colour, and shape, in that order).
The condition that a property has the same option across the three cards translates to the entry in the corresponding coordinate being the same across all three \(4\)-tuples. In the example above, all three cards are red. Therefore, the entries in the third coordinates of all three \(4\)-tuples are \(1\).
The condition that a property has all different options across the three cards translates to each of \(1\), \(2\), and \(3\) appearing in the corresponding coordinate in the three \(4\)-tuples. In the example above, all three cards have different shadings. Therefore, the entries \(1\), \(2\), and \(3\) appear in some order in the second coordinates of the three \(4\)-tuples.
Our goal now is to characterise exactly when a collection of three
numbers \(\{a,b,c\}\) is either \(\{1,1,1\}\), \(\{2,2,2\}\), \(\{3,3,3\}\), or \(\{1,2,3\}\). Note that in the last case,
\(a,b,c\) must be equal to \(1,2,3\) in some order (not necessarily
\(a = 1\), \(b = 2\), and \(c
= 3\)).
Suppose \(a,b,c\) are three integers,
each of which is either \(1\), \(2\), or \(3\). Then \(a + b
+ c\) is a multiple of three exactly when either \(a = b = c\) or \(a\), \(b\), and \(c\) are all distinct.
Let’s justify this claim. First, if \(a = b =
c\) then \(a + b +c\) is equal
to \(3\), \(6\), or \(9\). If \(a\), \(b\), and \(c\) are distinct, then \(a\), \(b\), and \(c\) are \(1\), \(2\), and \(3\) in some order. Therefore, \(a + b + c = 1 + 2 + 3 = 6\). Now suppose it
is not true that all three of \(a\),
\(b\), and \(c\) are equal or distinct. That means two
of the numbers are equal, and one is not. We can check the sum \(a + b + c\) in each of these cases. \[\begin{align*}
1 + 1 + 2 &= 4 \\
1 + 1 + 3 &= 5 \\
2 + 2 + 1 &= 5 \\
2 + 2 + 3 &= 7 \\
3 + 3 + 1 &= 7 \\
3 + 3 + 2 &= 8.\end{align*}\] In all of these cases, the sum
\(a + b + c\) is not a multiple of
\(3\), so the claim is true!
Great! Let’s harness this by adding up coordinates among tuples to check whether or not three \(4\)-tuples come from cards that form a set.
To make our lives a little easier, given two \(4\)-tuples, let’s add them together to create another \(4\)-tuple by simply adding up the coordinates. More precisely, given two \(4\)-tuples \(\textbf{v}= (a,b,c,d)\) and \(\textbf{w} = (p,q,r,s)\), we define \[\textbf{v} + \textbf{w} = (a + p, b + q, c + r, d + s).\] If you have experience with vectors, you may recognise that this is exactly how we add two vectors together. Using this definition of addition for \(4\)-tuples, we can repeatedly add as many \(4\)-tuples together as we like!
In particular, given three \(4\)-tuples we can add them together to get another \(4\)-tuple. The resulting \(4\)-tuple may include entries other than \(1\), \(2\), and \(3\), but that’s okay! By using the claim above, we know that the three \(4\)-tuples correspond to three cards that form a set exactly when the \(4\)-tuple resulting from adding them together has the property that every entry is a multiple of \(3\). Check for yourself that this works for the example set at the beginning of the solution to Problem \(5\).
More precisely, suppose \(\textbf{v}\), \(\textbf{w}\), and \(\textbf{u}\) are three \(4\)-tuples corresponding to SET! cards. Then \(\textbf{v} + \textbf{w} + \textbf{u}\) has every coordinate a multiple of \(3\) exactly when the cards corresponding to \(\textbf{v}\), \(\textbf{w}\), and \(\textbf{u}\) form a set. Choose your favourite set, and your favourite non-set, and check this for yourself!
We are now ready to attack the original question. First convert the \(81\) cards in a SET! deck into \(4\)-tuples. For each coordinate, the integers \(1\), \(2\), and \(3\) appear exactly \(27\) times each (this is because once the entry in a coordinate has been fixed, there are three different options for each of the remaining three coordinates, and \(3 \times 3 \times 3 = 27\)). Therefore, for each coordinate, the sum over all \(81\) entries in that coordinate is \(27\times 1 + 27 \times 2 + 27 \times 3 = 162\), which is a multiple of \(3\).
Considering the question at hand, we have collected \(26\) sets from a full SET! deck. We want to show that the remaining three cards also form a set. Let \(\textbf{w}_1,\textbf{w}_2,\textbf{w}_3\) be the three \(4\)-tuples corresponding to the remaining three cards. For the \(78\) cards used in the \(26\) sets, label the \(78\) \(4\)-tuples \(\textbf{v}_1,\textbf{v}_2,\ldots,\textbf{v}_{78}\) so that for every positive integer \(k \leq 26\), \(\textbf{v}_{3k-2},\textbf{v}_{3k-1}\), and \(\textbf{v}_{3k}\) form a set (ie, \(\textbf{v}_1, \textbf{v}_2, \textbf{v}_3\) form a set, \(\textbf{v}_4,\textbf{v}_5,\textbf{v}_6\) form a set and so on).
Then we know that for every positive integer \(k \leq 26\), \(\textbf{v}_{3k-2}+\textbf{v}_{3k-1}+\textbf{v}_{3k}\) is a \(4\)-tuple where every coordinate is a multiple of \(3\). Summing up all of the resulting \(4\)-tuples gives us that \[\textbf{v}_1 + \textbf{v}_2 + \cdots + \textbf{v}_{78} = (a,b,c,d)\] where \(a,b,c,d\) are all multiples of \(3\). We also know \[\textbf{v}_1 + \textbf{v}_2 + \cdots + \textbf{v}_{78} + \textbf{w}_1 + \textbf{w}_2 + \textbf{w}_3 = (162, 162, 162, 162).\] We can now conclude that \[\textbf{w}_1 + \textbf{w}_2 + \textbf{w}_3 = (162 - a, 162 - b, 162 - c, 162 - d).\] All of \(162\), \(a\), \(b\), \(c\), and \(d\) are all multiples of \(3\). Therefore, \(162 - a\), \(162 - b\), \(162 - c\), and \(162 - d\) are multiples of \(3\). We can finally conclude that \(\textbf{w}_1,\textbf{w}_2,\) and \(\textbf{w}_3\) are \(4\)-tuples corresponding to three cards that form a set.