Question

Let $X = \{a,b,c,d,e,f\}$ and $Y = \{7,8,9,10,11\}$ be two sets. Which one of the following is true?

• A. $\{(a,8),(b,7),(c,9),(d,10),(e,11)\}$ is one-to-one function from $X$ to $Y$
• B. $\{(a,7),(b,11),(c,8),(d,10),(e,9),(f,11)\}$ is one-to-one function from $X$ to $Y$
• C. $\{(a,7),(b,8),(c,9),(d,10),(e,11)\}$ is one-to-one function from $X$ to $Y$
• D. $\{(a,11),(b,10),(c,9),(d,8),(e,7),(f,9)\}$ is one-to-one function from $X$ to $Y$
• E. one-to-one function cannot be defined from $X$ to $Y$

Hint:

If number of elements in domain is greater than codomain, injective function is impossible (Pigeonhole Principle).

Answer 1

Answer: (E)

Concept:
• A function is one-to-one (injective) if distinct elements of domain have distinct images.
• Total number of elements in domain must be $\leq$ number of elements in codomain for injectivity.
• If $|X|>|Y|$, injective mapping is impossible (Pigeonhole Principle).

Step 1: Check cardinality of sets
\[ |X| = 6 \quad \text{and} \quad |Y| = 5 \]

Step 2: Apply Pigeonhole Principle
Since there are more elements in $X$ than in $Y$, at least two elements of $X$ must map to the same element in $Y$.
Hence, injective (one-to-one) mapping is not possible.

Step 3: Verify options individually
Option (A): Missing element $f$ $\Rightarrow$ not a function.
Option (B): $b$ and $f$ both map to $11$ $\Rightarrow$ not one-to-one.
Option (C): Missing element $f$ $\Rightarrow$ not a function.
Option (D): $c$ and $f$ both map to $9$ $\Rightarrow$ not one-to-one.
Option (E): Correct, as one-to-one function is not possible. Final Conclusion:
One-to-one function from $X$ to $Y$ cannot be defined.