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Question Number 57988 by peter frank last updated on 15/Apr/19
list all subset of   {2,4,6,7,8}
$${list}\:{all}\:{subset}\:{of}\: \\ $$$$\left\{\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{7},\mathrm{8}\right\} \\ $$
Commented by mr W last updated on 16/Apr/19
thanks sir!
$${thanks}\:{sir}! \\ $$
Commented by mr W last updated on 17/Apr/19
∅ (null set)⇒1=C_0 ^5   {2}{4}{6}{7}{8}⇒5=C_1 ^5   {24}{26}{27}{28}{46}{47}{48}{67}{67}{78}⇒10=C_2 ^5   {246}{247}{248}{267}{268}{278}{467}{468}{478}{678}⇒10=C_3 ^5   {2467}{2468}{2478}{2678}{4678}⇒5=C_4 ^5   {24678}⇒1=C_5 ^5   ⇒32=2^5   in general a set with n elements has  2^n  subsets.
$$\emptyset\:\left({null}\:{set}\right)\Rightarrow\mathrm{1}={C}_{\mathrm{0}} ^{\mathrm{5}} \\ $$$$\left\{\mathrm{2}\right\}\left\{\mathrm{4}\right\}\left\{\mathrm{6}\right\}\left\{\mathrm{7}\right\}\left\{\mathrm{8}\right\}\Rightarrow\mathrm{5}={C}_{\mathrm{1}} ^{\mathrm{5}} \\ $$$$\left\{\mathrm{24}\right\}\left\{\mathrm{26}\right\}\left\{\mathrm{27}\right\}\left\{\mathrm{28}\right\}\left\{\mathrm{46}\right\}\left\{\mathrm{47}\right\}\left\{\mathrm{48}\right\}\left\{\mathrm{67}\right\}\left\{\mathrm{67}\right\}\left\{\mathrm{78}\right\}\Rightarrow\mathrm{10}={C}_{\mathrm{2}} ^{\mathrm{5}} \\ $$$$\left\{\mathrm{246}\right\}\left\{\mathrm{247}\right\}\left\{\mathrm{248}\right\}\left\{\mathrm{267}\right\}\left\{\mathrm{268}\right\}\left\{\mathrm{278}\right\}\left\{\mathrm{467}\right\}\left\{\mathrm{468}\right\}\left\{\mathrm{478}\right\}\left\{\mathrm{678}\right\}\Rightarrow\mathrm{10}={C}_{\mathrm{3}} ^{\mathrm{5}} \\ $$$$\left\{\mathrm{2467}\right\}\left\{\mathrm{2468}\right\}\left\{\mathrm{2478}\right\}\left\{\mathrm{2678}\right\}\left\{\mathrm{4678}\right\}\Rightarrow\mathrm{5}={C}_{\mathrm{4}} ^{\mathrm{5}} \\ $$$$\left\{\mathrm{24678}\right\}\Rightarrow\mathrm{1}={C}_{\mathrm{5}} ^{\mathrm{5}} \\ $$$$\Rightarrow\mathrm{32}=\mathrm{2}^{\mathrm{5}} \\ $$$${in}\:{general}\:{a}\:{set}\:{with}\:{n}\:{elements}\:{has} \\ $$$$\mathrm{2}^{{n}} \:{subsets}. \\ $$
Commented by peter frank last updated on 15/Apr/19
thank you sir
$${thank}\:{you}\:{sir} \\ $$
Commented by Kunal12588 last updated on 16/Apr/19
i think  φ is also a subset of {2,4,6,7,8}  and every set is a subset of itself  ⇒{2,4,6,7,8}⊆{2,4,6,7,8}  so answer should be 2^5 =32
$${i}\:{think} \\ $$$$\phi\:{is}\:{also}\:{a}\:{subset}\:{of}\:\left\{\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{7},\mathrm{8}\right\} \\ $$$${and}\:{every}\:{set}\:{is}\:{a}\:{subset}\:{of}\:{itself} \\ $$$$\Rightarrow\left\{\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{7},\mathrm{8}\right\}\subseteq\left\{\mathrm{2},\mathrm{4},\mathrm{6},\mathrm{7},\mathrm{8}\right\} \\ $$$${so}\:{answer}\:{should}\:{be}\:\mathrm{2}^{\mathrm{5}} =\mathrm{32} \\ $$
Commented by problem solverd last updated on 16/Apr/19
but the question was to list  all subset
$$\boldsymbol{\mathrm{but}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{question}}\:\boldsymbol{\mathrm{was}}\:\boldsymbol{\mathrm{to}}\:\boldsymbol{\mathrm{list}} \\ $$$$\boldsymbol{\mathrm{all}}\:\boldsymbol{\mathrm{subset}} \\ $$

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