JEE MAIN - Mathematics (2021 - 22th July Evening Shift - No. 15)
Let A = {0, 1, 2, 3, 4, 5, 6, 7}. Then the number of bijective functions f : A $$\to$$ A such that f(1) + f(2) = 3 $$-$$ f(3) is equal to
Answer
720
Explanation
f(1) + f(2) = 3 $$-$$ f(3)
$$\Rightarrow$$ f(1) + f(2) = 3 + f(3) = 3
The only possibility is : 0 + 1 + 2 = 3
$$\Rightarrow$$ Elements 1, 2, 3 in the domain can be mapped with 0, 1, 2 only.
So number of bijective functions.
$$\left| \!{\underline {\, 3 \,}} \right. $$ $$\times$$ $$\left| \!{\underline {\, 5 \,}} \right. $$ = 720
$$\Rightarrow$$ f(1) + f(2) = 3 + f(3) = 3
The only possibility is : 0 + 1 + 2 = 3
$$\Rightarrow$$ Elements 1, 2, 3 in the domain can be mapped with 0, 1, 2 only.
So number of bijective functions.
$$\left| \!{\underline {\, 3 \,}} \right. $$ $$\times$$ $$\left| \!{\underline {\, 5 \,}} \right. $$ = 720
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