Key Concepts and Formulas

• One-One Function (Injective Function): A function $f: A \to B$ is one-one if distinct elements in the domain $A$ map to distinct elements in the codomain $B$.
• Number of One-One Functions: The number of one-one functions from a set of size $m$ to a set of size $n$ (where $m \le n$) is given by the permutation formula $P(n, m) = \frac{n!}{(n-m)!} = n \times (n-1) \times \dots \times (n-m+1)$. If $m > n$, the number of one-one functions is 0.
• Cardinality of Cartesian Product: For finite sets $A$ and $B$, the cardinality of their Cartesian product is $|A \times B| = |A| \times |B|$.

Step-by-Step Solution

Step 1: Understand the given sets and the definitions of x and y. We are given a set $A$ with $|A| = 3$ elements and a set $B$ with $|B| = 5$ elements. $x$ is the total number of one-one functions from set $A$ to set $B$. $y$ is the total number of one-one functions from set $A$ to the set $A \times B$.

Step 2: Calculate x, the number of one-one functions from A to B. Here, the domain is set $A$ with $m = |A| = 3$ elements, and the codomain is set $B$ with $n = |B| = 5$ elements. Since $m \le n$ ($3 \le 5$), one-one functions can exist. The number of one-one functions $x$ is given by $P(n, m) = P(5, 3)$. $x = P(5, 3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = 5 \times 4 \times 3$ $x = 60$

Step 3: Calculate the cardinality of the set A $\times$ B. The set $A \times B$ is the Cartesian product of sets $A$ and $B$. Its cardinality is the product of the cardinalities of $A$ and $B$. $|A \times B| = |A| \times |B| = 3 \times 5$ $|A \times B| = 15$

Step 4: Calculate y, the number of one-one functions from A to A $\times$ B. Here, the domain is set $A$ with $m = |A| = 3$ elements, and the codomain is set $A \times B$ with $n = |A \times B| = 15$ elements. Since $m \le n$ ($3 \le 15$), one-one functions can exist. The number of one-one functions $y$ is given by $P(n, m) = P(15, 3)$. $y = P(15, 3) = \frac{15!}{(15-3)!} = \frac{15!}{12!} = 15 \times 14 \times 13$ $y = 2730$

Step 5: Establish the relationship between x and y. We have $x = 60$ and $y = 2730$. We need to find a relationship between them that matches one of the options. Let's consider the ratio $\frac{y}{x}$. $\frac{y}{x} = \frac{2730}{60}$ We can simplify this fraction: $\frac{y}{x} = \frac{273}{6}$ Both numerator and denominator are divisible by 3: $\frac{y}{x} = \frac{91}{2}$ Multiplying both sides by 2 gives: $2y = 91x$

Common Mistakes & Tips

• Incorrectly calculating the size of the Cartesian product: Always remember $|A \times B| = |A| \times |B|$.
• Confusing permutations with combinations: For one-one functions, the order matters as distinct domain elements must map to distinct codomain elements, so permutations ($P(n,m)$) are used, not combinations.
• Forgetting the condition $m \le n$: If the number of elements in the domain is greater than the number of elements in the codomain, the number of one-one functions is zero.

Summary

We calculated $x$ as the number of one-one functions from a set of 3 elements to a set of 5 elements, yielding $x = P(5, 3) = 60$. We then calculated $y$ as the number of one-one functions from a set of 3 elements to a set of $3 \times 5 = 15$ elements, yielding $y = P(15, 3) = 2730$. By finding the ratio $\frac{y}{x} = \frac{2730}{60} = \frac{91}{2}$, we derived the relationship $2y = 91x$.

The final answer is \boxed{2y = 91x} which corresponds to option (A).