1. We need the number of one-one (injective) maps $f:A\to B$ where $A=\{1,3,7,9,11\},\qquad B=\{2,4,5,7,8,10,12\}$ and the condition is $f(1)+f(3)=14.$

Since $f$ is one-one, we must have $f(1)\neq f(3)$.

2. First find all ordered pairs $(f(1),f(3))$ from $B$ such that $f(1)+f(3)=14.$

Elements of $B$ are $\{2,4,5,7,8,10,12\}$. Possible sums to $14$ are:

• $2+12=14$
• $4+10=14$

Also $7+7=14$, but this is not allowed because injective map requires distinct images for $1$ and $3$.

So the valid ordered pairs are: $(2,12),(12,2),(4,10),(10,4)$ Hence, number of choices for $(f(1),f(3))$ is $4.$

3. After fixing $f(1)$ and $f(3)$, there remain $7-2=5$ elements of $B$ available for the remaining elements $7,9,11\in A$.

Since the map must remain one-one, we assign distinct images to these 3 elements from the remaining 5 elements.

Number of ways: $^5P_3=5\cdot 4\cdot 3=60.$

4. Therefore total number of injective maps is $4\times 60=240.$

5. Checking options:

• A: $120$
• B: $180$
• C: $240$ ✅
• D: $480$

So the correct answer is Option C.