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Don't panic bro . We are here to help .

this is just an example of bipirated graph. . one set contains printer and one contain 8 computers. no edge can be between the same set. as it is said to guarentee atleast 4 access different printers. so take one computer connect with all printer . 4 cables. now take one more , and connect with 4 . similarly connect 4 computer with all the 4 printers. 4*4 = 16 cables are consumed and we have satisfied the condition. now just connect remaining 4 computer with the minimum no of connection  i.e 1 .which will lead to 4 . so 16+4 =20 .
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There are eight computers and four printers.One set contains 8 computers $C_1,C_2...C_8$and one set contains 4 printers $P_1,P_2,P_3,P_4$.It is given that four computers $(C_1,C_2,C_3,C_4)$ can directly access 4 printers.It means $(C_1,C_2,C_3,C_4)$ should be connect to the $(P_1,P_2,P_3,P_4)$.

Total cables required to connect 4 computers to 4 different printers = $4\times 4=16$

Now remaining 4 computers $(C_5, C_6, C_7, C_8)$ can connect with a minimum number of cables to the printer which is 1.For each remaining computers, one cable required to connect printers.

Hence, Least number of cables required to connect 8 computers to 4 printers=16+4=20

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