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

	Answered by sniper237 last updated on 18/Feb/25
	$Let E={f:A→B} , ∣E∣=3^m E\S={f∈E, ∣f(A)∣=1 or 2} ∣E\S∣= C_3 ^1 .1^m +C_3 ^2 .(2^m −1^m −1^m )=3.2^m −3 So ∣S∣=3^m −(3.2^m −3)$
	$L e t E = {f : A \to B}, ∣ E ∣= 3^{m}$ $E ∖ S = {f \in E, ∣ f (A) ∣= 1 o r 2}$ $∣ E ∖ S ∣= C_{3}^{1} . 1^{m} + C_{3}^{2} . (2^{m} - 1^{m} - 1^{m}) = 3 . 2^{m} - 3$ $S o ∣ S ∣= 3^{m} - (3 . 2^{m} - 3)$
	Commented by mehdee7396 last updated on 18/Feb/25

	$n o t e : 3^{m} - (3 \times 2^{m} - 3) \neq 3^{m} - 3 \times 2^{m} - 3$
	Commented by mnjuly1970 last updated on 18/Feb/25

	$t h a n k s a l o t . . .$
	Answered by mehdee7396 last updated on 19/Feb/25
	$let A={a_i : 1≤i≤m} & B={b_1 ,b_2 ,b_3 } F={f:A→B∣f(a)=b ;a∈A , b∈B} f_1 ={f:A→B∣f(a_i )≠b_1 } f_2 ={f:A→B∣f(a_2 )≠b_2 } f_3 ={f:A→B∣f(a_i )≠b_3 } ⇒∣F∣=3^m & ∣f_i ∣=2^m & ∣f_i ∩f_j ∣=1 & ∣f_1 ∩f_2 ∩f_3 ∣=0 ⇒∣S∣=∣F∣−Σ∣f_i ∣+Σ∣f_i ∩f_j ∣−∣f_1 ∩f_2 ∩f_3 ∣ =3^m −3×2^m +3$
	$l e t A = {a_{i} : 1 ⩽ i ⩽ m} & B = {b_{1}, b_{2}, b_{3}}$ $F = {f : A \to B ∣ f (a) = b; a \in A, b \in B}$ $f_{1} = {f : A \to B ∣ f (a_{i}) \neq b_{1}}$ $f_{2} = {f : A \to B ∣ f (a_{2}) \neq b_{2}}$ $f_{3} = {f : A \to B ∣ f (a_{i}) \neq b_{3}}$ $\Rightarrow∣ F ∣= 3^{m} & ∣ f_{i} ∣= 2^{m} & ∣ f_{i} \cap f_{j} ∣= 1 & ∣ f_{1} \cap f_{2} \cap f_{3} ∣= 0$ $\Rightarrow∣ S ∣=∣ F ∣ - Σ ∣ f_{i} ∣ + Σ ∣ f_{i} \cap f_{j} ∣ - ∣ f_{1} \cap f_{2} \cap f_{3} ∣$ $= 3^{m} - 3 \times 2^{m} + 3$
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