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Question Number 34063 by rahul 19 last updated on 30/Apr/18

Let A= { 1,2,3,4 } . Number of functions f:A→A satisfying f(f(x))=x ∀x∈A, is ?

$$\boldsymbol{\mathrm{L}}\mathrm{et}\:\mathrm{A}=\:\left\{\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\:\right\}\:.\:\mathrm{N}{umber}\:\mathrm{of}\:\mathrm{functions} \\ $$$$\mathrm{f}:\mathrm{A}\rightarrow{A}\:\mathrm{satisfying}\:\mathrm{f}\left(\mathrm{f}\left({x}\right)\right)={x}\:\forall{x}\in\mathrm{A},\:\mathrm{is}\:? \\ $$

Commented by rahul 19 last updated on 30/Apr/18

Ans. is 13....

$${Ans}.\:{is}\:\mathrm{13}.... \\ $$

Commented by rahul 19 last updated on 30/Apr/18

does this gives 13?

$${does}\:{this}\:{gives}\:\mathrm{13}? \\ $$

Commented by Rasheed.Sindhi last updated on 30/Apr/18

For example: {(1,2),(2,1),(3,4),(4,3)} f(1)=2, f( f(1) )=f(2)=1 f( f(2) )=f(1)=2 a,b,c,d∈A , a,b,c,d are different {(a,a),(b,b),(c,c),(d,d)} {(a,a),(b,b),(c,d),(d,c)} {(a,b),(b,a),(c,d),(d,c)} Continue

$$\mathrm{For}\:\mathrm{example}: \\ $$$$\:\:\:\:\:\:\left\{\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{2},\mathrm{1}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{4},\mathrm{3}\right)\right\} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\mathrm{2},\:\mathrm{f}\left(\:\mathrm{f}\left(\mathrm{1}\right)\:\right)=\mathrm{f}\left(\mathrm{2}\right)=\mathrm{1} \\ $$$$\mathrm{f}\left(\:\mathrm{f}\left(\mathrm{2}\right)\:\right)=\mathrm{f}\left(\mathrm{1}\right)=\mathrm{2} \\ $$$$ \\ $$$$\:\:\:\:\:\:{a},{b},{c},{d}\in\mathrm{A}\:,\:{a},{b},{c},{d}\:\mathrm{are}\:\mathrm{different} \\ $$$$\:\:\:\:\:\:\:\left\{\left({a},{a}\right),\left({b},{b}\right),\left({c},{c}\right),\left({d},{d}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\left\{\left({a},{a}\right),\left({b},{b}\right),\left({c},{d}\right),\left({d},{c}\right)\right\} \\ $$$$\:\:\:\:\:\:\left\{\left({a},{b}\right),\left({b},{a}\right),\left({c},{d}\right),\left({d},{c}\right)\right\} \\ $$$$ \\ $$$$\mathrm{Continue} \\ $$

Answered by Rasheed.Sindhi last updated on 30/Apr/18

Domain(f)=A, the function have 4 orderd pairs, the pair of f is either of (a,a) type or if it is (a,b) with a≠b then the f also contains (b,a) because of the condition f(f(x))=x ∀x∈A. The ordered pair containing 1 as abscissa may be (1,1),(1,2),(1,3),(1,4) (1,1):(2,2),(2,3),(2,4) [(2,1) is impossible −−−−−− (1,1),(2,2),(3,3),(4,4) (1,1),(2,2),(3,4),(4,3) (1,1),(2,3),(3,2),(4,4) (1,1),(2,4),(3,3),(4,2) −−−−− (1,2),(2,1),(3,4),(4,3) (1,2),(2,1),(3,3),(4,4) (1,3),(2,4),(3,1),(4,2) (1,3),(2,2),(3,1),(4,4) (1,4)(2,3)(3,2),(4,1) (1,4)(2,2)(3,3),(4,1)

$$\mathrm{Domain}\left(\mathrm{f}\right)=\mathrm{A},\:\mathrm{the}\:\mathrm{function}\:\mathrm{have}\:\mathrm{4}\:\mathrm{orderd} \\ $$$$\mathrm{pairs},\:\mathrm{the}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{f}\:\mathrm{is}\:\mathrm{either}\:\mathrm{of}\:\left({a},{a}\right)\:\mathrm{type} \\ $$$$\mathrm{or}\:\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\left({a},{b}\right)\:\mathrm{with}\:{a}\neq\mathrm{b}\:\mathrm{then}\:\mathrm{the}\:\mathrm{f}\:\mathrm{also} \\ $$$$\mathrm{contains}\:\left({b},{a}\right)\:\mathrm{because}\:\mathrm{of}\:\mathrm{the}\:\mathrm{condition} \\ $$$$\mathrm{f}\left(\mathrm{f}\left({x}\right)\right)={x}\:\forall{x}\in\mathrm{A}. \\ $$$$\:\:\:\:\:\:\:\mathrm{The}\:\mathrm{ordered}\:\mathrm{pair}\:\mathrm{containing}\:\mathrm{1}\:\mathrm{as}\:\mathrm{abscissa} \\ $$$$\mathrm{may}\:\mathrm{be}\:\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{1},\mathrm{3}\right),\left(\mathrm{1},\mathrm{4}\right) \\ $$$$\:\left(\mathrm{1},\mathrm{1}\right):\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{2},\mathrm{3}\right),\left(\mathrm{2},\mathrm{4}\right)\:\left[\left(\mathrm{2},\mathrm{1}\right)\:\mathrm{is}\:\mathrm{impossible}\right. \\ $$$$−−−−−− \\ $$$$\:\:\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{3},\mathrm{3}\right),\left(\mathrm{4},\mathrm{4}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{4},\mathrm{3}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{2},\mathrm{3}\right),\left(\mathrm{3},\mathrm{2}\right),\left(\mathrm{4},\mathrm{4}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{2},\mathrm{4}\right),\left(\mathrm{3},\mathrm{3}\right),\left(\mathrm{4},\mathrm{2}\right) \\ $$$$−−−−− \\ $$$$\:\:\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{2},\mathrm{1}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{4},\mathrm{3}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{2},\mathrm{1}\right),\left(\mathrm{3},\mathrm{3}\right),\left(\mathrm{4},\mathrm{4}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{3}\right),\left(\mathrm{2},\mathrm{4}\right),\left(\mathrm{3},\mathrm{1}\right),\left(\mathrm{4},\mathrm{2}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{3}\right),\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{3},\mathrm{1}\right),\left(\mathrm{4},\mathrm{4}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{4}\right)\left(\mathrm{2},\mathrm{3}\right)\left(\mathrm{3},\mathrm{2}\right),\left(\mathrm{4},\mathrm{1}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{4}\right)\left(\mathrm{2},\mathrm{2}\right)\left(\mathrm{3},\mathrm{3}\right),\left(\mathrm{4},\mathrm{1}\right) \\ $$$$\: \\ $$

Commented by Rasheed.Sindhi last updated on 30/Apr/18

I think the answer should be 10.

$$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\mathrm{10}. \\ $$

Commented by rahul 19 last updated on 30/Apr/18

Thank You Sir !

$$\mathscr{T}{hank}\:\mathscr{Y}{ou}\:\mathscr{S}{ir}\:! \\ $$

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