Key Concepts and Formulas

• Natural Numbers ($\mathbb{N}$): The set of positive integers, $\{1, 2, 3, \dots\}$.
• 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 $B$. That is, if $x_1, x_2 \in A$ and $x_1 \neq x_2$, then $f(x_1) \neq f(x_2)$.
• Counting One-One Functions: If $A$ and $B$ are finite sets with $|A| = m$ and $|B| = n$, the number of one-one functions from $A$ to $B$ is given by the permutation formula $P(n, m) = \frac{n!}{(n-m)!}$, provided $n \ge m$. If $n < m$, the number of one-one functions is 0. If $n=m$, the number is $n!$.

Step-by-Step Solution

Step 1: Determine the elements of set A The set $A$ is defined as $A = \{(x, y) : 2x + 3y = 23, x, y \in \mathbb{N}\}$. We need to find all pairs of natural numbers $(x, y)$ that satisfy the equation $2x + 3y = 23$. Since $x, y \in \mathbb{N}$, we know $x \ge 1$ and $y \ge 1$. From $2x = 23 - 3y$, we must have $23 - 3y > 0$, which implies $3y < 23$, so $y < \frac{23}{3}$. Since $y$ is a natural number, $y$ can take values from $\{1, 2, 3, 4, 5, 6, 7\}$. From $3y = 23 - 2x$, we must have $23 - 2x > 0$, which implies $2x < 23$, so $x < \frac{23}{2}$. Since $x$ is a natural number, $x$ can take values from $\{1, 2, \dots, 11\}$.

We can systematically check values of $y$ from 1 to 7 to find corresponding integer values of $x$:

• If $y=1$: $2x + 3(1) = 23 \Rightarrow 2x = 20 \Rightarrow x = 10$. Since $x=10 \in \mathbb{N}$, $(10, 1) \in A$.
• If $y=2$: $2x + 3(2) = 23 \Rightarrow 2x = 17 \Rightarrow x = 17/2$. Not an integer.
• If $y=3$: $2x + 3(3) = 23 \Rightarrow 2x = 14 \Rightarrow x = 7$. Since $x=7 \in \mathbb{N}$, $(7, 3) \in A$.
• If $y=4$: $2x + 3(4) = 23 \Rightarrow 2x = 11 \Rightarrow x = 11/2$. Not an integer.
• If $y=5$: $2x + 3(5) = 23 \Rightarrow 2x = 8 \Rightarrow x = 4$. Since $x=4 \in \mathbb{N}$, $(4, 5) \in A$.
• If $y=6$: $2x + 3(6) = 23 \Rightarrow 2x = 5 \Rightarrow x = 5/2$. Not an integer.
• If $y=7$: $2x + 3(7) = 23 \Rightarrow 2x = 2 \Rightarrow x = 1$. Since $x=1 \in \mathbb{N}$, $(1, 7) \in A$.

Thus, the set $A$ contains the following ordered pairs: $A = \{(1, 7), (4, 5), (7, 3), (10, 1)\}$ The cardinality of set $A$ is $|A| = 4$.

Step 2: Determine the elements of set B The set $B$ is defined as $B = \{x : (x, y) \in A\}$. This means set $B$ consists of all the first components (the $x$-values) of the ordered pairs in set $A$. From the elements of $A$:

• For $(1, 7)$, the $x$-value is 1.
• For $(4, 5)$, the $x$-value is 4.
• For $(7, 3)$, the $x$-value is 7.
• For $(10, 1)$, the $x$-value is 10.

Therefore, the set $B$ is: $B = \{1, 4, 7, 10\}$ The cardinality of set $B$ is $|B| = 4$.

Step 3: Calculate the number of one-one functions from A to B We need to find the number of one-one functions from set $A$ to set $B$. We have $|A| = 4$ and $|B| = 4$. Since the cardinality of the domain ($|A|$) is equal to the cardinality of the codomain ($|B|$), a one-one function exists. The number of one-one functions from a set of size $m$ to a set of size $n$ (where $n \ge m$) is given by $P(n, m)$. In this case, $m = |A| = 4$ and $n = |B| = 4$. The number of one-one functions is $P(4, 4) = \frac{4!}{(4-4)!} = \frac{4!}{0!} = \frac{24}{1} = 24$.

Common Mistakes & Tips

• Definition of Natural Numbers: Ensure you are using the correct definition of natural numbers ($\mathbb{N} = \{1, 2, 3, \dots\}$ for JEE context) to avoid including or excluding 0 incorrectly.
• Diophantine Equation Solutions: When solving linear Diophantine equations with constraints (like $x, y \in \mathbb{N}$), systematically check all possible values for one variable within the derived bounds to guarantee finding all solutions.
• Cardinality Comparison: Always verify that the cardinality of the codomain is greater than or equal to the cardinality of the domain ($|B| \ge |A|$) before calculating the number of one-one functions. If $|A| > |B|$, the number of one-one functions is 0.

Summary We first identified the elements of set $A$ by solving the linear Diophantine equation $2x + 3y = 23$ for natural numbers $x$ and $y$, finding that $A = \{(1, 7), (4, 5), (7, 3), (10, 1)\}$, so $|A|=4$. Next, we determined set $B$ by collecting the $x$-components of the pairs in $A$, yielding $B = \{1, 4, 7, 10\}$, so $|B|=4$. Since we are looking for one-one functions from $A$ to $B$, and $|A|=|B|=4$, the number of such functions is given by $P(4, 4)$, which equals $4! = 24$.

The final answer is $\boxed{24}$.