Let A = {1, 2, 3, 4, 5, 6}. The number of one | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(NONE) Given $f: A 	o A$ is a one-one function where $A = \{1, 2, 3, 4, 5, 6\}$.We have the condition $f(2) + f(3) = 5$ and $f(3) \le 4$.The possible

Vedclass · Accepted Answer

$120$

Vedclass · Answer

(NONE) Given $f: A 	o A$ is a one-one function where $A = \{1, 2, 3, 4, 5, 6\}$.We have the condition $f(2) + f(3) = 5$ and $f(3) \le 4$.The possible pairs $(f(2), f(3))$ are $(1, 4), (4, 1), (2, 3), (3, 2)$.Since $f$ is one-one, $f(1), f(2), f(3)$ must be distinct.Case $1$: $(f(2), f(3)) = (1, 4)$. Then $f(1) \in A \setminus \{1, 4\} = \{2, 3, 5, 6\}$. Given $f(1) \ge 3$, so $f(1) \in \{3, 5, 6\}$ ($3$ choices). Remaining $3$ elements can be mapped in $3! = 6$ ways. Total $= 3 	imes 6 = 18$.Case $2$: $(f(2), f(3)) = (4, 1)$. Then $f(1) \in A \setminus \{4, 1\} = \{2, 3, 5, 6\}$. Given $f(1) \ge 3$, so $f(1) \in \{3, 5, 6\}$ ($3$ choices). Remaining $3$ elements can be mapped in $3! = 6$ ways. Total $= 3 	imes 6 = 18$.Case $3$: $(f(2), f(3)) = (2, 3)$. Then $f(1) \in A \setminus \{2, 3\} = \{1, 4, 5, 6\}$. Given $f(1) \ge 3$, so $f(1) \in \{4, 5, 6\}$ ($3$ choices). Remaining $3$ elements can be mapped in $3! = 6$ ways. Total $= 3 	imes 6 = 18$.Case $4$: $(f(2), f(3)) = (3, 2)$. Then $f(1) \in A \setminus \{3, 2\} = \{1, 4, 5, 6\}$. Given $f(1) \ge 3$, so $f(1) \in \{4, 5, 6\}$ ($3$ choices). Remaining $3$ elements can be mapped in $3! = 6$ ways. Total $= 3 	imes 6 = 18$.Total number of functions $= 18 + 18 + 18 + 18 = 72$.

Let $A = \{1, 2, 3, 4, 5, 6\}$. The number of one-one functions $f: A \to A$ such that $f(1) \ge 3, f(3) \le 4$ and $f(2) + f(3) = 5$, is ————

Similar Questions

Which of the following is an even function?

The function $f(x) = \sin (\log (x + \sqrt {x^2 + 1}))$ is

If $f : R \to R$ is defined by $f(x) = 2x + \cos x$,then $f$ is

If $f: R \to R$,then $f(x) = |x|$ is

Show that the function $f: N \rightarrow N$,given by $f(1)=f(2)=1$ and $f(x)=x-1$ for every $x>2$,is onto but not one-one.

Vedclass Test Series

Exam Paper Generator

Online Exam Module