Animated Solution for Mathematics - Sets and Relations: Comprehension Passage

A Journey into Counting

Welcome, fellow explorers of mathematics! Today, we are going to peel back the layers of a beautiful problem involving sets, relations, and the elegant logic of counting.

Often, when we see a problem involving relations on a set $S=\{1,2,3,4,5,6\}$, our minds might jump to complex definitions. But let's pause and visualize the core of the problem: a $6\times 6$ grid.

The Forbidden Zone

Imagine a grid where the x-axis represents the first element $a$ and the y-axis represents the second element $b$. A relation $R$ is simply a collection of points on this grid.

The condition $|a-b|\ge 2$ is our filter. It tells us that $a$ and $b$ cannot be the same (the main diagonal, where $|a-b|=0$) and they cannot be neighbors (the adjacent diagonals, where $|a-b|=1$).

If you take a moment to sketch this, you will see that we are removing $6$ points from the main diagonal and $5+5=10$ points from the adjacent diagonals. That is $16$ forbidden points out of $36$.

This leaves us with exactly $20$ valid points in our 'playground'. This set of $20$ points is the only source from which we can pick our elements for the relation $R$.

Thus, the number of such relations $n(X)$ is simply the number of ways to choose any subset of these $20$ points, which is $2^{20}$. This confirms our value $m=20$.

The Mystery of the Range

Now, let's tackle $Y$, the set of relations where the range has exactly one element. If the range is a single element $y$, then every pair in our relation must be of the form $(x,y)$.

This means for a fixed $y$, we need to find $x$ values such that $|x-y|\ge 2$. Let's test $y=3$. The valid $x$ values are those where $|x-3|\ge 2$, which means $x\le 1$ or $x\ge 5$.

That gives us $x\in \{1,5,6\}$. That is only $3$ values! No matter which $y$ you pick, you will never find enough $x$ values to satisfy a relation of a specific size if the range is restricted to one element. Therefore, $n(Y)=0$.

The Function Constraint

Finally, we arrive at $Z$, the set of relations that are functions. A function $f:S\to S$ requires that for every $x\in S$, there is exactly one $f(x)\in S$ such that $|x-f(x)|\ge 2$.

We simply count the valid choices for each $x$:

- For $x=1$, $f(1)\in \{3,4,5,6\}$ ($4$ choices) - For $x=2$, $f(2)\in \{4,5,6\}$ ($3$ choices) - For $x=3$, $f(3)\in \{1,5,6\}$ ($3$ choices) - For $x=4$, $f(4)\in \{1,2,6\}$ ($3$ choices) - For $x=5$, $f(5)\in \{1,2,3\}$ ($3$ choices) - For $x=6$, $f(6)\in \{1,2,3,4\}$ ($4$ choices)

Multiplying these together, we get:

Since $1296=36^2$, we find that $k=36$. The beauty of this problem lies in how the constraints, which initially seem restrictive, guide us toward a clean, perfect square.

Keep exploring, and remember: every complex problem is just a collection of simple, logical steps waiting to be uncovered!