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Question Number 216734 by mnjuly1970 last updated on 17/Feb/25

Answered by sniper237 last updated on 18/Feb/25

Let E={f:A→B} , ∣E∣=3^m E\S={f∈E, ∣f(A)∣=1 or 2} ∣E\S∣= C_3 ^1 .1^m +C_3 ^2 .(2^m −1^m −1^m )=3.2^m −3 So ∣S∣=3^m −(3.2^m −3)

$${Let}\:{E}=\left\{{f}:{A}\rightarrow{B}\right\}\:\:,\:\mid{E}\mid=\mathrm{3}^{{m}} \: \\ $$$${E}\backslash{S}=\left\{{f}\in{E},\:\:\mid{f}\left({A}\right)\mid=\mathrm{1}\:{or}\:\mathrm{2}\right\} \\ $$$$\mid{E}\backslash{S}\mid=\:{C}_{\mathrm{3}} ^{\mathrm{1}} .\mathrm{1}^{{m}} +{C}_{\mathrm{3}} ^{\mathrm{2}} .\left(\mathrm{2}^{{m}} −\mathrm{1}^{{m}} −\mathrm{1}^{{m}} \right)=\mathrm{3}.\mathrm{2}^{{m}} −\mathrm{3} \\ $$$${So}\:\:\mid{S}\mid=\mathrm{3}^{{m}} −\left(\mathrm{3}.\mathrm{2}^{{m}} −\mathrm{3}\right) \\ $$

Commented by mehdee7396 last updated on 18/Feb/25

note : 3^m −(3×2^m −3)≠3^m −3×2^m −3

$${note}\::\:\mathrm{3}^{{m}} −\left(\mathrm{3}×\mathrm{2}^{{m}} −\mathrm{3}\right)\neq\mathrm{3}^{{m}} −\mathrm{3}×\mathrm{2}^{{m}} −\mathrm{3} \\ $$

Commented by mnjuly1970 last updated on 18/Feb/25

thanks alot...

$$\:{thanks}\:{alot}... \\ $$

Answered by mehdee7396 last updated on 19/Feb/25

let A={a_i : 1≤i≤m} & B={b_1 ,b_2 ,b_3 } F={f:A→B∣f(a)=b ;a∈A , b∈B} f_1 ={f:A→B∣f(a_i )≠b_1 } f_2 ={f:A→B∣f(a_2 )≠b_2 } f_3 ={f:A→B∣f(a_i )≠b_3 } ⇒∣F∣=3^m & ∣f_i ∣=2^m & ∣f_i ∩f_j ∣=1 & ∣f_1 ∩f_2 ∩f_3 ∣=0 ⇒∣S∣=∣F∣−Σ∣f_i ∣+Σ∣f_i ∩f_j ∣−∣f_1 ∩f_2 ∩f_3 ∣ =3^m −3×2^m +3

$${let}\:\:{A}=\left\{{a}_{{i}} :\:\mathrm{1}\leqslant{i}\leqslant{m}\right\}\:\&\:\:{B}=\left\{{b}_{\mathrm{1}} ,{b}_{\mathrm{2}} ,{b}_{\mathrm{3}} \right\} \\ $$$${F}=\left\{{f}:{A}\rightarrow{B}\mid{f}\left({a}\right)={b}\:\:;{a}\in{A}\:,\:{b}\in{B}\right\} \\ $$$${f}_{\mathrm{1}} =\left\{{f}:{A}\rightarrow{B}\mid{f}\left({a}_{{i}} \right)\neq{b}_{\mathrm{1}} \right\} \\ $$$${f}_{\mathrm{2}} =\left\{{f}:{A}\rightarrow{B}\mid{f}\left({a}_{\mathrm{2}} \right)\neq{b}_{\mathrm{2}} \right\} \\ $$$${f}_{\mathrm{3}} =\left\{{f}:{A}\rightarrow{B}\mid{f}\left({a}_{{i}} \right)\neq{b}_{\mathrm{3}} \right\} \\ $$$$\Rightarrow\mid{F}\mid=\mathrm{3}^{{m}} \:\:\&\:\:\mid{f}_{{i}} \mid=\mathrm{2}^{{m}} \:\&\:\:\mid{f}_{{i}} \cap{f}_{{j}} \mid=\mathrm{1}\:\:\&\:\:\mid{f}_{\mathrm{1}} \cap{f}_{\mathrm{2}} \cap{f}_{\mathrm{3}} \mid=\mathrm{0} \\ $$$$\Rightarrow\mid{S}\mid=\mid{F}\mid−\Sigma\mid{f}_{{i}} \mid+\Sigma\mid{f}_{{i}} \cap{f}_{{j}} \mid−\mid{f}_{\mathrm{1}} \cap{f}_{\mathrm{2}} \cap{f}_{\mathrm{3}} \mid \\ $$$$=\mathrm{3}^{{m}} −\mathrm{3}×\mathrm{2}^{{m}} +\mathrm{3} \\ $$$$ \\ $$

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