Let A - 0- 1- 2- 3- 4- 5- 6- 7- Then the number of bijective functions f-A- A such that f1 - f2 - 3 - f3 is equal to

F(1) + f(2) = 3 f(3) ⇒ f(1) + f(2) +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. ∴ Total number ...

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Standard X

Mathematics

Difference of Two Sets

Let A = 0, 1,...

Question

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

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Solution

Clearly $f (1), f (2)$ and $f (3)$ are the permutations of $0, 1, 2$ ; and $f (0), f (4), f (5), f (6)$ and $f (7)$ are the permutations of $3, 4, 5, 6$ and $7$ .

Total number of bijective functions = $5! 3! = 720$

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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

Clearly f(1),f(2) and f(3) are the permutations of 0,1,2; and f(0),f(4),f(5),f(6) and f(7) are the permutations of 3,4,5,6 and 7. Total number of bijective functions =5!3!=720

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

Clearly $f (1), f (2)$ and $f (3)$ are the permutations of $0, 1, 2$ ; and $f (0), f (4), f (5), f (6)$ and $f (7)$ are the permutations of $3, 4, 5, 6$ and $7$ .

Total number of bijective functions = $5! 3! = 720$