"Let $A=\{1,3,7,9,11\}$ and $B=\{2,4,5,7,8,10,12\}$. Then the total number of one-one maps $f: A \rightarrow B$, such that $f(1)+f(3)=14$, is :"

Question

Accepted Answer

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$f(1)+f(3)=14$

Case I

$$\begin{aligned} & f(1)=2, f(3)=12 \\ & f(1)=12, f(3)=2 \end{aligned}$$

Total one-one function

$$\begin{aligned} & =2 \times 5 \times 4 \times 3 \\ & =120 \end{aligned}$$

Case II

$$\begin{aligned} & f(1)=4, f(3)=10 \\ & f(1)=10, f(3)=4 \end{aligned}$$

Total one-one function

$$\begin{aligned} & =2 \times 5 \times 4 \times 3 \\ & =120 \\ & \text { Total cases }=120+120=240 \end{aligned}$$

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Let $A=\{1,3,7,9,11\}$ and $B=\{2,4,5,7,8,10,12\}$. Then the total number of one-one maps $f: A \rightarrow B$, such that $f(1)+f(3)=14$, is :