Let A = {2, 3, 6, 7} and B = {4, 5, 6, 8}. Let R be a relation defined on A × B by (a1, b1) R (a2, b

Q: Why do we rearrange the equation? It looks fine as it is.

It is all about perspective. The original form is a constraint on four variables simultaneously, which is hard to visualize. The rearranged form, $a_1 - b_1 = b_2 - a_2$, separates the variables into two independent 'difference' blocks. This turns a 4D problem into a 1D frequency matching problem, which is much easier to solve.

Q: Why is $m_k = n_{-k}$?

Great observation! $m_k$ is defined as the frequency of $b - a = k$. If we multiply both sides by $-1$, we get $a - b = -k$. Since $n_{-k}$ is the frequency of $a - b = -k$, they are mathematically identical. It is just a matter of symmetry.

Q: Is there a risk of double counting here?

Only if you are not systematic. By using the frequency table method, we partition the solution space into disjoint sets based on the value of $k$. Since each pair $(a, b)$ produces exactly one value of $k$, every valid pair is counted exactly once. It is a very robust method!

Question

Let $A = \{2, 3, 6, 7\}$ and $B = \{4, 5, 6, 8\}$. Let $R$ be a relation defined on $A 	imes B$ by $(a_1, b_1) R (a_2, b_2)$ if and only if $a_1 + a_2 = b_1 + b_2$. Then the number of elements in $R$ is _________

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem asks us to find the number of elements in a relation $R$ defined on $A 	imes B$ such that $a_1 + a_2 = b_1 + b_2$. At first glance, this involves four variables and appears daunting.

However, mathematics is about finding the hidden structure beneath the surface. By shifting our perspective, we can simplify the constraint significantly.

### The Power of Rearrangement

Consider the equation:
$$a_1 + a_2 = b_1 + b_2$$

By rearranging the terms to group variables by their set membership, we obtain:
$$a_1 - b_1 = b_2 - a_2$$

Let this common difference be $k$. We now seek pairs such that $a_1 - b_1 = k$ and $a_2 - b_2 = -k$. This transformation decouples the variables and turns a four-variable constraint into a frequency matching problem.

### The Grid of Possibilities

Given sets $A = \{2, 3, 6, 7\}$ and $B = \{4, 5, 6, 8\}$, we calculate the difference $d = a - b$ for every possible pair $(a, b)$.

We define $n_k$ as the number of times a difference $k$ occurs. By systematically populating the grid, we identify the frequency of each difference $k$:

*   For $k = -6$: $(2-8)$, $n_{-6} = 1$
*   For $k = -5$: $(2-7)$ is not possible, but $(3-8)$, $n_{-5} = 1$
*   For $k = -4$: $(2-6), (3-7)$ is not possible, $(6-8)$ is not possible, $n_{-4} = 1$
*   For $k = -3$: $(2-5), (3-6)$, $n_{-3} = 2$
*   For $k = -2$: $(2-4), (3-5), (6-8)$, $n_{-2} = 3$
*   For $k = -1$: $(6-7)$ is not possible, $(7-8), (6-7)$ is not possible, $(3-4)$, $n_{-1} = 2$
*   For $k = 0$: $(6-6)$, $n_0 = 1$
*   For $k = 1$: $(6-5), (7-6)$, $n_1 = 2$
*   For $k = 2$: $(6-4), (7-5)$, $n_2 = 2$
*   For $k = 3$: $(7-4)$, $n_3 = 1$

### The Frequency Dance

The total number of valid relations is the sum of the products of these frequencies, where we pair a difference $k$ with its negative counterpart $-k$:
$$	ext{Total} = \sum n_k 	imes n_{-k}$$

For every pair $(a_1, b_1)$ that yields a difference $k$, we must pair it with a $(a_2, b_2)$ that yields a difference $-k$. This ensures that $a_1 - b_1 = -(a_2 - b_2)$, which satisfies the original condition $a_1 + a_2 = b_1 + b_2$.

### Final Calculation

We sum the products for all possible values of $k$:

*   $k = \pm 3$: $n_3 	imes n_{-3} = 1 	imes 2 = 2$
*   $k = \pm 2$: $n_2 	imes n_{-2} = 2 	imes 3 = 6$
*   $k = \pm 1$: $n_1 	imes n_{-1} = 2 	imes 2 = 4$
*   $k = 0$: $n_0 	imes n_0 = 1 	imes 1 = 1$
*   $k = \mp 1$: $n_{-1} 	imes n_1 = 2 	imes 2 = 4$
*   $k = \mp 2$: $n_{-2} 	imes n_2 = 3 	imes 2 = 6$
*   $k = \mp 3$: $n_{-3} 	imes n_3 = 2 	imes 1 = 2$

Summing these values:
$$2 + 6 + 4 + 1 + 4 + 6 + 2 = 25$$

The total number of elements in the relation is **25**.

Animated Solution for Mathematics - Sets and Relations: Let $A = {2, 3, 6, 7}$ and $B = {4, 5, 6, 8}$ . Let $R$ be a relation defined on $A \times B$ by $(a_{1}, b_{1}) R (a_{2}, b_{2})$ if and only if $a_{1} + a_{2} = b_{1} + b_{2}$ . Then the number of elements in $R$ is _________

Enter Numerical Value:

Visualized Solution

Understanding the Sets

Analyzing the Condition

Calculating $a - b$ Values

Defining Frequencies $n_{k}$ and $m_{k}$

Frequencies for $k = - 3$ and $k = - 2$

Frequencies for $k = - 1$ and $k = 0$

Frequencies for $k = 1, 2, 3$

Final Summation

The Sigma Insight: Types of Relations

Analyzing the Setup

The Power of Rearrangement

The Grid of Possibilities

The Frequency Dance

Final Calculation

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