Let A= 0,1,2,3,4,5,6,7 . Then the number of bijective functions f: A arrow A such that f(1)+f(2)=3-f(3) is equal to

Question

Let A= 0,1,2,3,4,5,6,7 . Then the number of bijective functions f: A arrow A such that f(1)+f(2)=3-f(3) is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

f(1)+f(2)=3-f(3) ⇒ f(1)+f(2)=3+f(3)=3 The only possibility is: 0+1+2=3 ⇒ Elements 1,2,3 in the domain can be mapped with 0,1,2 only. So number of bijective functions. =3 ! × 5 !=720

Q. 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

$f (1) + f (2) = 3 - f (3)$
$\Rightarrow f (1) + f (2) = 3 + f (3) = 3$
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.
$= 3! \times 5! = 720$

Q. Let A={0,1,2,3,4,5,6,7}. Then the number of bijective functions f:A→A such that f(1)+f(2)=3−f(3) is equal to

f(1)+f(2)=3−f(3) ⇒f(1)+f(2)=3+f(3)=3 The only possibility is: 0+1+2=3 ⇒ Elements 1,2,3 in the domain can be mapped with 0,1,2 only. So number of bijective functions. =3!×5!=720

Q. 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

$f (1) + f (2) = 3 - f (3)$
$\Rightarrow f (1) + f (2) = 3 + f (3) = 3$
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.
$= 3! \times 5! = 720$