Let [A = 1,2,3,....,10 ] and [f:A A] be defined as [f(k) = pmatrix k + 1 & if,k,is,odd k &…

Question

Let $$A = \{ 1,2,3,....,10\} $$ and $$f:A \to A$$ be defined as

$$f(k) = \left\{ {\begin{pmatrix} {k + 1} & {if\,k\,is\,odd}  \
   k & {if\,k\,is\,even}  \ \end{pmatrix} } \right.$$

Then the number of possible functions $$g:A \to A$$ such that $gof = f$ is :

Accepted Answer

f(1) = 2

f(2) = 2

f(3) = 4

f(4) = 4

f(5) = 6

f(6) = 6

f(7) = 8

f(8) = 8

f(9) = 10

f(10) = 10

$$ \therefore $$ f(1) = f(2) = 2

f(3) = f(4) = 4

f(5) = f(6) = 6

f(7) = f(8) = 8

f(9) = f(10) = 10

Given, g(f(x)) = f(x)

when x = 1, g(f(1)) = f(1) $$ \Rightarrow $$ g(2) = 2

when, x = 2, g(f(2)) = f(2) $$ \Rightarrow $$ g(2) = 2

$$ \therefore $$ x = 1, 2, g(2) = 2

Similarly, at x = 3, 4, g(4) = 4

at x = 5, 6, g(6) = 6

at x = 7, 8, g(8) = 8

at x = 9, 10, g(10) = 10

Here, you can see for even terms mapping is fixed. But far odd terms 1, 3, 5, 7, 9 we can map to any one of the 10 elements.

$$ \therefore $$ For 1, number of functions = 10

For 3, number of functions = 10

$$\eqalign{
  \& .  \cr
  \& .  \cr
  \& .  \cr

\&  \cr} $$
for 9, number of functions = 10

$$ \therefore $$ Total number of functions = 10 $$\times$$ 10 $$\times$$ 10 $$\times$$ 10 $$\times$$ 10 = 105

Solutions · 1