Even elements in the domain are $2$ and $4$.
Even elements in the codomain are $2,4,6,8$.
The two even elements of the domain must be mapped to even elements of the codomain injectively.
Number of ways:
${}^4P_2=4\times 3=12$
Now, $2$ elements of the codomain are already used, so $6$ elements are left.
The odd elements of the domain are $1$ and $3$.
They can be mapped injectively to the remaining $6$ elements.
Number of ways:
${}^6P_2=6\times 5=30$
So,
$n=12\times 30=360$
Now,
$360=2^3\times 3^2\times 5^1$
So,
$a=3,\ b=2,\ c=1$
Therefore,
$a+b+c=3+2+1=6$
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Online Test Series, Information About Examination,
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and More.