Question

The least number of cables required to connect 8 computers to 4 printers to guarantee that 4 computers can directly access 4 different printers.

At any given time 4 computers should be able to simultaneously access 4 printers.

I assumed this was simple bipartite graph k(8,4) , 32 connections.That seems to be wrong.

Answer 1

(20) Answer Iam getting is 20..

Since each 4 comp need direct connected with each printer.... so 16 connection + now remaining 4 computer, each connected to 4 different printers, so 4  connections=20 connections. 

c1-> p1,p2,p3,p4

c2-> p1,p2,p3,p4

c3-> p1,p2,p3,p4

c4-> p1,p2,p3,p4

c5->p1

c6->p2

c7->p3

c8->p4

Now, any pick of 4 computers will have a direct connection to all the 4 printers. 

Comments

"remaining 4 computer, each connected to different printers " rt?

Question is quite ambiguous. guess your interpretation is what the question setter had in mind. 

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yes Arjun, rest 4 computers can be connected to any one printer each

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confusing statement...but good explanation

Answer 2

refer to this:http://math.stackexchange.com/questions/351948/the-basic-of-the-count

Comments

The answer on math stack  is 28, but actually it's given as 20.Still trying to figure it out.

Answer 3

Answer is 16. Consider following connections:

1:  1,2,3

2:  2,3,4

3:  1,2,4

4:  1,3,4  

Other 4 computers connected to 1,2,3,4 printers : one_one connection.

Comments

Question is a bit ambiguous- 16 is the answer if we can choose how the connections can be made. 28 is the answer if connections can be made anyway and not controlled by us.

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You're right , it is rather ambiguous.

If we think like that ,

Answer could very well be 29 :

You see, 8*3 : All computers connected to 3 printers ,

Then 5 more minimum to ensure that any 4 computers selected would have at least one computer which is connected to all 4 printers.