To determine which subsets of \( X \times Y \) are functions from \( X \) to \( Y \), we need to recall the definition of a function. A relation from set \( X \) to set \( Y \) is a function if:
1. Every element in \( X \) (the domain) must be associated with at least one element in \( Y \).
2. Each element in \( X \) must be associated with a unique element in \( Y \) (no element in \( X \) can map to multiple elements in \( Y \)).
Given:
- \( X = \{a, b, c, d, e\} \)
- \( Y = \{p, q, r, s, t\} \)
Let's analyze each option step-by-step.
### Step 1: Analyze the first option
**Option 1:** \( \{(a, r), (b, r), (c, s), (d, r), (e, p)\} \)
- For \( a \): maps to \( r \) (unique)
- For \( b \): maps to \( r \) (not unique, since \( b \) and \( a \) both map to \( r \))
- For \( c \): maps to \( s \) (unique)
- For \( d \): maps to \( r \) (not unique)
- For \( e \): maps to \( p \) (unique)
Since \( b \) and \( d \) both map to \( r \), this option does not satisfy the uniqueness condition. Thus, **Option 1 is not a function.**
### Step 2: Analyze the second option
**Option 2:** \( \{(a, r), (b, r), (c, t), (d, q), (e, s)\} \)
- For \( a \): maps to \( r \) (unique)
- For \( b \): maps to \( r \) (not unique)
- For \( c \): maps to \( t \) (unique)
- For \( d \): maps to \( q \) (unique)
- For \( e \): maps to \( s \) (unique)
Again, \( a \) and \( b \) both map to \( r \), violating the uniqueness condition. Thus, **Option 2 is not a function.**
### Step 3: Analyze the third option
**Option 3:** \( \{(a, p), (b, t), (c, r), (d, s), (e, q)\} \)
- For \( a \): maps to \( p \) (unique)
- For \( b \): maps to \( t \) (unique)
- For \( c \): maps to \( r \) (unique)
- For \( d \): maps to \( s \) (unique)
- For \( e \): maps to \( q \) (unique)
Each element in \( X \) maps to a unique element in \( Y \), satisfying both conditions. Thus, **Option 3 is a function.**
### Step 4: Analyze the fourth option
**Option 4:** \( \{(a, r), (b, r), (c, r), (d, r), (e, r)\} \)
- For \( a \): maps to \( r \) (unique)
- For \( b \): maps to \( r \) (not unique)
- For \( c \): maps to \( r \) (not unique)
- For \( d \): maps to \( r \) (not unique)
- For \( e \): maps to \( r \) (not unique)
All elements in \( X \) map to the same element \( r \) in \( Y \). While each \( X \) maps to a single \( Y \), the uniqueness condition is violated since multiple elements in \( X \) map to the same element in \( Y \). Thus, **Option 4 is not a function.**
### Conclusion
The only subset that satisfies the conditions of being a function from \( X \) to \( Y \) is **Option 3**.
---
