We are given that:
\( A = \{(x, y) : 2x + 3y = 23, \, x, y \in \mathbb{N} \} \)
\( B = \{x : (x, y) \in A\} \)
We need to find the number of one-to-one functions from \( A \) to \( B \).
Step 1: Find the Elements of Set A
We are given the equation \( 2x + 3y = 23 \), where \( x \) and \( y \) are natural numbers (\( \mathbb{N} \)).
To solve for \( y \) in terms of \( x \), we rearrange the equation:
\( 3y = 23 - 2x \Rightarrow y = \frac{23 - 2x}{3} \)
For \( y \) to be a natural number, \( 23 - 2x \) must be divisible by 3. Thus, we need to solve the congruence:
\( 23 - 2x \equiv 0 \pmod{3} \)
Simplifying:
\( 23 \equiv 2 \pmod{3} \) and \( 2x \equiv 2 \pmod{3} \)
\( x \equiv 1 \pmod{3} \)
Thus, \( x \) must be of the form \( x = 3k + 1 \) for some integer \( k \). Now, let’s substitute values of \( x \) into the equation \( 2x + 3y = 23 \) and solve for \( y \).
For \( x = 1 \):
\( 2(1) + 3y = 23 \Rightarrow 2 + 3y = 23 \Rightarrow 3y = 21 \Rightarrow y = 7 \)
Thus, \( (x, y) = (1, 7) \).
For \( x = 4 \):
\( 2(4) + 3y = 23 \Rightarrow 8 + 3y = 23 \Rightarrow 3y = 15 \Rightarrow y = 5 \)
Thus, \( (x, y) = (4, 5) \).
For \( x = 7 \):
\( 2(7) + 3y = 23 \Rightarrow 14 + 3y = 23 \Rightarrow 3y = 9 \Rightarrow y = 3 \)
Thus, \( (x, y) = (7, 3) \).
For \( x = 10 \):
\( 2(10) + 3y = 23 \Rightarrow 20 + 3y = 23 \Rightarrow 3y = 3 \Rightarrow y = 1 \)
Thus, \( (x, y) = (10, 1) \).
So, the elements of set \( A \) are:
\( A = \{(1, 7), (4, 5), (7, 3), (10, 1)\} \)
Step 2: Define Set \( B \)
Set \( B = \{x : (x, y) \in A\} \). Thus, \( B = \{1, 4, 7, 10\} \).
Step 3: Find the Number of One-to-One Functions
The number of one-to-one functions from \( A \) to \( B \) is the number of ways to assign each element of \( A \) to a unique element of \( B \). Since both sets \( A \) and \( B \) contain 4 elements, the number of one-to-one functions is simply the number of permutations of 4 elements, which is:
\( 4! = 24 \)
Thus, the number of one-to-one functions from \( A \) to \( B \) is:
24
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