Let A={1,3,7,9,11} and B={2,4,5,7,8,10,12} . Then total number of one-one maps mathrm{f}: mathrm{A}

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1.Relation and Function

hard

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

A

$180$

B

$120$

C

$480$

D

$240$

(JEE MAIN-2024)

Solution

$ A=\{1,3,7,9,11\} $

$ B=\{2,4,5,7,8,10,12\} $

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

$ \text { (i) } 2+12 $

$ \text { (ii) } 4+10 $

$ 2 \times(2 \times 5 \times 4 \times 3)=240$

Standard 12

Mathematics

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