Let R be a relation defined on the set {1, 2, 3, 4} × {1, 2, 3, 4} by R = {((a, b), (c, d)) : 2a + 3

Q: Wait, why can't I just iterate through all $256$ possible pairs? Isn't that safer?

While brute force is technically possible, it is the enemy of speed and accuracy in JEE Advanced. By calculating frequencies, you reduce the cognitive load and minimize the chance of a silly arithmetic error. Trust the structure! The frequency method is a hallmark of a disciplined mathematical mind.

Q: What if I miscount the frequencies in the table? Is there a way to check?

Great question! Always check your range. For $2a+3b$, the minimum is $2(1)+3(1)=5$ and the maximum is $2(4)+3(4)=20$. If your table doesn't cover this range, you've missed something. Stay systematic by iterating through the variables one by one, just as we did in the steps.

Q: Why do we multiply the frequencies? I thought we were supposed to add them.

Think of the Fundamental Counting Principle. For a specific value $k$, every pair $(a,b)$ that results in $k$ can be paired with every pair $(c,d)$ that also results in $k$. If there are $n$ ways to get $k$ on the left and $m$ ways on the right, the total combinations for that $k$ is $n \times m$. We only add the results *after* we have calculated the combinations for each unique $k$.

Question

Let $R$ be a relation defined on the set $\{1, 2, 3, 4\} 	imes \{1, 2, 3, 4\}$ by $R = \{((a, b), (c, d)) : 2a + 3b = 3c + 4d\}$. Then the number of elements in $R$ is

Options:

(A) (1) 18

(B) (2) 6

(C) (3) 15

(D) (4) 12

Sigma Percentile · Accepted Answer

### Analyzing the Setup

The problem requires us to find the number of elements in a relation $R$ defined on the set $A = \{1, 2, 3, 4\}$. The relation is governed by the equation:

$$2a + 3b = 3c + 4d$$

To solve this, we define two functions: $L(a, b) = 2a + 3b$ and $R(c, d) = 3c + 4d$, where $a, b, c, d \in \{1, 2, 3, 4\}$. We seek the number of quadruplets $((a, b), (c, d))$ such that $L(a, b) = R(c, d) = k$.

### Phase 1: Mapping the Left-Hand World

We calculate all possible values of $L(a, b) = 2a + 3b$ for $a, b \in \{1, 2, 3, 4\}$ and record their frequencies:

*   For $b=1$: $L \in \{5, 7, 9, 11\}$
*   For $b=2$: $L \in \{8, 10, 12, 14\}$
*   For $b=3$: $L \in \{11, 13, 15, 17\}$
*   For $b=4$: $L \in \{14, 16, 18, 20\}$

The frequency distribution for $L$ is:
$5:1, 7:1, 8:1, 9:1, 10:1, 11:2, 12:1, 13:1, 14:2, 15:1, 16:1, 17:1, 18:1, 20:1$.

### Phase 2: Mapping the Right-Hand World

Next, we calculate all possible values of $R(c, d) = 3c + 4d$ for $c, d \in \{1, 2, 3, 4\}$ and record their frequencies:

*   For $d=1$: $R \in \{7, 10, 13, 16\}$
*   For $d=2$: $R \in \{11, 14, 17, 20\}$
*   For $d=3$: $R \in \{15, 18, 21, 24\}$
*   For $d=4$: $R \in \{19, 22, 25, 28\}$

The frequency distribution for $R$ is:
$7:1, 10:1, 11:1, 13:1, 14:1, 15:1, 16:1, 17:1, 18:1, 19:1, 20:1, 21:1, 22:1, 24:1, 25:1, 28:1$.

### Phase 3: The Intersection and the Grand Tally

We identify the common values $k$ and multiply their respective frequencies from both worlds using the **Fundamental Counting Principle**. The total number of elements is the sum of these products:

$$
\begin{aligned}
k=7: & \quad 1 	imes 1 = 1 \
k=10: & \quad 1 	imes 1 = 1 \
k=11: & \quad 2 	imes 1 = 2 \
k=13: & \quad 1 	imes 1 = 1 \
k=14: & \quad 2 	imes 1 = 2 \
k=15: & \quad 1 	imes 1 = 1 \
k=16: & \quad 1 	imes 1 = 1 \
k=17: & \quad 1 	imes 1 = 1 \
k=18: & \quad 1 	imes 1 = 1 \
k=20: & \quad 1 	imes 1 = 1
\end{aligned}
$$

Summing these values, we get $1 + 1 + 2 + 1 + 2 + 1 + 1 + 1 + 1 + 1 = 12$.

The total number of elements in the relation $R$ is **12**.

Animated Solution for Mathematics - Sets and Relations: Let $R$ be a relation defined on the set ${1, 2, 3, 4} \times {1, 2, 3, 4}$ by $R = {((a, b), (c, d)) : 2 a + 3 b = 3 c + 4 d}$ . Then the number of elements in $R$ is

Select Answer:

Visualized Solution

Understanding the Relation $R$

The Strategy: Finding Common Values

Tabulating LHS: $2 a + 3 b$

Tabulating RHS: $3 c + 4 d$

Identifying Common Values $k$

Counting for $k = 7$ and $k = 10$

Counting for $k = 11$ and $k = 13$

Counting for $k = 14$ and $k = 15$

Counting for $k = 16, 17, 18, 20$

Final Calculation

The Sigma Insight: Domain and Range of a Relation

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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Select Answer:

Visualized Solution

Understanding the Relation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

The Strategy: Finding Common Values

Tabulating LHS: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2a+3b

Tabulating RHS: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3c+4d

Identifying Common Values .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k

Counting for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=7 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=10

Counting for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=11 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=13

Counting for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=14 and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=15

Counting for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k=16,17,18,20