Key Concepts and Formulas

• One-One Map (Injective Function): A function $f: A \rightarrow B$ is one-one if distinct elements in the domain $A$ map to distinct elements in the codomain $B$. That is, if $a_1 \neq a_2$ for $a_1, a_2 \in A$, then $f(a_1) \neq f(a_2)$.
• Number of One-One Maps: If $|A|=m$ and $|B|=n$ with $n \ge m$, the total number of one-one maps from $A$ to $B$ is given by the number of permutations of $n$ items taken $m$ at a time, denoted as $P(n,m)$ or $^nP_m$, where $P(n,m) = \frac{n!}{(n-m)!}$.

Step-by-Step Solution

Step 1: Understand the Given Sets and the Constraint We are given the domain $A = \{1, 3, 7, 9, 11\}$ and the codomain $B = \{2, 4, 5, 7, 8, 10, 12\}$. The size of the domain is $|A| = 5$. The size of the codomain is $|B| = 7$. We need to find the number of one-one maps $f: A \rightarrow B$ such that $f(1) + f(3) = 14$.

Step 2: Identify Valid Pairs for $f(1)$ and $f(3)$ from Set $B$ Since $f$ is a one-one map, $f(1)$ and $f(3)$ must be distinct elements from set $B$. We need to find pairs of distinct elements $(x, y)$ from $B$ such that $x+y=14$. Let's examine elements of $B$: $\{2, 4, 5, 7, 8, 10, 12\}$. \begin{itemize} \item If $f(1)=2$, then $f(3)=14-2=12$. Both $2, 12 \in B$. This is a valid pair. \item If $f(1)=4$, then $f(3)=14-4=10$. Both $4, 10 \in B$. This is a valid pair. \item If $f(1)=5$, then $f(3)=14-5=9$. $9 \notin B$. This is not a valid pair. \item If $f(1)=7$, then $f(3)=14-7=7$. For a one-one map, $f(1) \neq f(3)$, so this is not valid. \item If $f(1)=8$, then $f(3)=14-8=6$. $6 \notin B$. This is not a valid pair. \item If $f(1)=10$, then $f(3)=14-10=4$. Both $10, 4 \in B$. This is a valid pair. \item If $f(1)=12$, then $f(3)=14-12=2$. Both $12, 2 \in B$. This is a valid pair. \end{itemize} The possible pairs of values for $\{f(1), f(3)\}$ are $\{2, 12\}$ and $\{4, 10\}$.

Step 3: Calculate the Number of One-One Maps for Each Valid Pair Case We have two mutually exclusive cases based on the pairs identified in Step 2.

Case 1: $\{f(1), f(3)\} = \{2, 12\}$ In this case, $f(1)$ and $f(3)$ must be assigned the values $2$ and $12$. There are $2! = 2$ ways to assign these values: \begin{enumerate} \item $f(1)=2$ and $f(3)=12$ \item $f(1)=12$ and $f(3)=2$ \end{enumerate} After assigning values to $f(1)$ and $f(3)$, two distinct elements from $B$ are used. The remaining elements in $A$ are $\{7, 9, 11\}$, which are $5 - 2 = 3$ elements. The remaining elements in $B$ are $7 - 2 = 5$ elements. We need to map the remaining 3 elements of $A$ to the remaining 5 elements of $B$ in a one-one manner. The number of ways to do this is given by the permutation formula $P(5,3)$: $P(5,3) = \frac{5!}{(5-3)!} = \frac{5!}{2!} = 5 \times 4 \times 3 = 60$ So, for Case 1, the total number of one-one maps is the product of the number of ways to assign $f(1), f(3)$ and the number of ways to map the remaining elements: Number of maps for Case 1 = $2 \times P(5,3) = 2 \times 60 = 120$.

Case 2: $\{f(1), f(3)\} = \{4, 10\}$ Similarly, $f(1)$ and $f(3)$ must be assigned the values $4$ and $10$. There are $2! = 2$ ways to assign these values: \begin{enumerate} \item $f(1)=4$ and $f(3)=10$ \item $f(1)=10$ and $f(3)=4$ \end{enumerate} Again, two distinct elements from $B$ are used. The remaining 3 elements of $A$ need to be mapped to the remaining 5 elements of $B$ one-one. The number of ways is $P(5,3)$: $P(5,3) = 5 \times 4 \times 3 = 60$ So, for Case 2, the total number of one-one maps is: Number of maps for Case 2 = $2 \times P(5,3) = 2 \times 60 = 120$.

Step 4: Calculate the Total Number of One-One Maps Since Case 1 and Case 2 are mutually exclusive, we add the number of maps from each case to find the total number of one-one maps satisfying the condition: Total number of one-one maps = Number of maps for Case 1 + Number of maps for Case 2 Total number of one-one maps = $120 + 120 = 240$.

Common Mistakes & Tips

• Distinctness for One-One: Always ensure that $f(a_1) \neq f(a_2)$ when $a_1 \neq a_2$. This was crucial in ruling out $f(1)=7, f(3)=7$.
• Order Matters for Assignments: When assigning specific values to specific domain elements (like $f(1)$ and $f(3)$), the order matters. For each pair of values, there are $2!$ ways to assign them.
• Set Membership: Verify that all mapped values are indeed elements of the codomain $B$.

Summary We identified pairs of distinct elements in $B$ that sum to 14, finding two such pairs: $\{2, 12\}$ and $\{4, 10\}$. For each pair, there are 2 ways to assign these values to $f(1)$ and $f(3)$. The remaining 3 elements of $A$ must be mapped one-one to the remaining 5 elements of $B$, which can be done in $P(5,3)=60$ ways. Summing the possibilities for both pairs gives a total of $2 \times 60 + 2 \times 60 = 240$ one-one maps.

The final answer is $\boxed{240}$, which corresponds to option (C).