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

$$\boldsymbol{\mathrm{L}}\mathrm{et}\:\mathrm{A}=\:\left\{\:\mathrm{1},\mathrm{2},\mathrm{3},\mathrm{4}\:\right\}\:.\:\mathrm{N}{umber}\:\mathrm{of}\:\mathrm{functions} \\ $$$$\mathrm{f}:\mathrm{A}\rightarrow{A}\:\mathrm{satisfying}\:\mathrm{f}\left(\mathrm{f}\left({x}\right)\right)={x}\:\forall{x}\in\mathrm{A},\:\mathrm{is}\:? \\ $$
Commented by rahul 19 last updated on 30/Apr/18

$${Ans}.\:{is}\:\mathrm{13}…. \\ $$
Commented by rahul 19 last updated on 30/Apr/18

$${does}\:{this}\:{gives}\:\mathrm{13}? \\ $$
Commented by Rasheed.Sindhi last updated on 30/Apr/18

$$\mathrm{For}\:\mathrm{example}: \\ $$$$\:\:\:\:\:\:\left\{\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{2},\mathrm{1}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{4},\mathrm{3}\right)\right\} \\ $$$$\mathrm{f}\left(\mathrm{1}\right)=\mathrm{2},\:\mathrm{f}\left(\:\mathrm{f}\left(\mathrm{1}\right)\:\right)=\mathrm{f}\left(\mathrm{2}\right)=\mathrm{1} \\ $$$$\mathrm{f}\left(\:\mathrm{f}\left(\mathrm{2}\right)\:\right)=\mathrm{f}\left(\mathrm{1}\right)=\mathrm{2} \\ $$$$ \\ $$$$\:\:\:\:\:\:{a},{b},{c},{d}\in\mathrm{A}\:,\:{a},{b},{c},{d}\:\mathrm{are}\:\mathrm{different} \\ $$$$\:\:\:\:\:\:\:\left\{\left({a},{a}\right),\left({b},{b}\right),\left({c},{c}\right),\left({d},{d}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\left\{\left({a},{a}\right),\left({b},{b}\right),\left({c},{d}\right),\left({d},{c}\right)\right\} \\ $$$$\:\:\:\:\:\:\left\{\left({a},{b}\right),\left({b},{a}\right),\left({c},{d}\right),\left({d},{c}\right)\right\} \\ $$$$ \\ $$$$\mathrm{Continue} \\ $$
Answered by Rasheed.Sindhi last updated on 30/Apr/18

$$\mathrm{Domain}\left(\mathrm{f}\right)=\mathrm{A},\:\mathrm{the}\:\mathrm{function}\:\mathrm{have}\:\mathrm{4}\:\mathrm{orderd} \\ $$$$\mathrm{pairs},\:\mathrm{the}\:\mathrm{pair}\:\mathrm{of}\:\mathrm{f}\:\mathrm{is}\:\mathrm{either}\:\mathrm{of}\:\left({a},{a}\right)\:\mathrm{type} \\ $$$$\mathrm{or}\:\mathrm{if}\:\mathrm{it}\:\mathrm{is}\:\left({a},{b}\right)\:\mathrm{with}\:{a}\neq\mathrm{b}\:\mathrm{then}\:\mathrm{the}\:\mathrm{f}\:\mathrm{also} \\ $$$$\mathrm{contains}\:\left({b},{a}\right)\:\mathrm{because}\:\mathrm{of}\:\mathrm{the}\:\mathrm{condition} \\ $$$$\mathrm{f}\left(\mathrm{f}\left({x}\right)\right)={x}\:\forall{x}\in\mathrm{A}. \\ $$$$\:\:\:\:\:\:\:\mathrm{The}\:\mathrm{ordered}\:\mathrm{pair}\:\mathrm{containing}\:\mathrm{1}\:\mathrm{as}\:\mathrm{abscissa} \\ $$$$\mathrm{may}\:\mathrm{be}\:\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{1},\mathrm{3}\right),\left(\mathrm{1},\mathrm{4}\right) \\ $$$$\:\left(\mathrm{1},\mathrm{1}\right):\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{2},\mathrm{3}\right),\left(\mathrm{2},\mathrm{4}\right)\:\left[\left(\mathrm{2},\mathrm{1}\right)\:\mathrm{is}\:\mathrm{impossible}\right. \\ $$$$−−−−−− \\ $$$$\:\:\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{3},\mathrm{3}\right),\left(\mathrm{4},\mathrm{4}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{4},\mathrm{3}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{2},\mathrm{3}\right),\left(\mathrm{3},\mathrm{2}\right),\left(\mathrm{4},\mathrm{4}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{1}\right),\left(\mathrm{2},\mathrm{4}\right),\left(\mathrm{3},\mathrm{3}\right),\left(\mathrm{4},\mathrm{2}\right) \\ $$$$−−−−− \\ $$$$\:\:\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{2},\mathrm{1}\right),\left(\mathrm{3},\mathrm{4}\right),\left(\mathrm{4},\mathrm{3}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{2}\right),\left(\mathrm{2},\mathrm{1}\right),\left(\mathrm{3},\mathrm{3}\right),\left(\mathrm{4},\mathrm{4}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{3}\right),\left(\mathrm{2},\mathrm{4}\right),\left(\mathrm{3},\mathrm{1}\right),\left(\mathrm{4},\mathrm{2}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{3}\right),\left(\mathrm{2},\mathrm{2}\right),\left(\mathrm{3},\mathrm{1}\right),\left(\mathrm{4},\mathrm{4}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{4}\right)\left(\mathrm{2},\mathrm{3}\right)\left(\mathrm{3},\mathrm{2}\right),\left(\mathrm{4},\mathrm{1}\right) \\ $$$$\:\:\left(\mathrm{1},\mathrm{4}\right)\left(\mathrm{2},\mathrm{2}\right)\left(\mathrm{3},\mathrm{3}\right),\left(\mathrm{4},\mathrm{1}\right) \\ $$$$\: \\ $$
Commented by Rasheed.Sindhi last updated on 30/Apr/18

$$\mathrm{I}\:\mathrm{think}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{should}\:\mathrm{be}\:\mathrm{10}. \\ $$
Commented by rahul 19 last updated on 30/Apr/18

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