+89 This is question from assignment book
A computer operation subroutine detects 19 computers and 23 peripheral graphics devices on a network. Each peripheral must be connected to a computer and each computer must be connected to a peripheral. In how many ways can the subroutine link the connections?
This has answer in back of book but I not kno how to do this one
+1036 Geno’s solution applies if the setup instructions restrict a computer to one graphic device at a time and allows a computer to be disconnected from all graphic devices.
The set up in the question requires a computer connection to at least one peripheral graphics device and allows for more than one connection. The peripheral graphics device require at least one connection to a computer and allows for more than one connection. Because of the bidirectional nature of this communication, if a computer is connected to a peripheral, the peripheral is also connected to the computer. This may seem obvious but the connection requirements would not work if this were not so.
This kind of software is usually installed in a print server. The computers poll the peripherals to find the status and the peripherals broadcast their status to the computers. Similar software controls network connections. The connection protocols are relatively trivial, while the algorithms that optimize the routing hops through nodes are complex and intricate. The solution to this is found using this formula. N = number of computers and k = number of peripheral graphics devices.
$$\displaystyle \sum \limits_{i=0}^{\textcolor[rgb]{1,0,0}{N-1}} *(-1)^i*\binom{N}{i}* (N-i)^k \hspace{15pt}| \hspace{15pt} \Text {N=19 \; k=23} \\\$$
Wolfram Alpha: scripted link
Script
sum (binom(19,i)*(-1)^i*(19-i)^23) from i=0 to 18
This result equals the one presented in your book.
+23246 If there can be only one computer connect to one peripheral graphics device (leaving 4 graphics devices unconnected) then:
the first computer can be connected to 23 devices;
the second computer can be connected to 22 devices;
the third computer can be connected to 21 devices;
etc;
the nineteenth computer can be connected to 5 devices;
so 23 x 22 x 21 x ... x 5 or 23! / 4!
+89 Thank you for answer
This 23!/4! = 1 077 167 364 120 207 360 000
The book show this as answer
9 705 062 517 250 244 272 128 000
This seem very to big. Is book wrong?
+118608 19 computers and 23 peripheral graphics devices
I don't get this question.
The connections are not 1 to 1 since there are more peripheral devices than computers.
So some computers are connected to more than 1 device.
What is the maximum number of divices that a computer can connect to?
Can a peripheral connect to more than one computer?
Se my confusion. 
The answer could be really enormous depending on what the question really means .
+1036 Geno’s solution applies if the setup instructions restrict a computer to one graphic device at a time and allows a computer to be disconnected from all graphic devices.
The set up in the question requires a computer connection to at least one peripheral graphics device and allows for more than one connection. The peripheral graphics device require at least one connection to a computer and allows for more than one connection. Because of the bidirectional nature of this communication, if a computer is connected to a peripheral, the peripheral is also connected to the computer. This may seem obvious but the connection requirements would not work if this were not so.
This kind of software is usually installed in a print server. The computers poll the peripherals to find the status and the peripherals broadcast their status to the computers. Similar software controls network connections. The connection protocols are relatively trivial, while the algorithms that optimize the routing hops through nodes are complex and intricate. The solution to this is found using this formula. N = number of computers and k = number of peripheral graphics devices.
$$\displaystyle \sum \limits_{i=0}^{\textcolor[rgb]{1,0,0}{N-1}} *(-1)^i*\binom{N}{i}* (N-i)^k \hspace{15pt}| \hspace{15pt} \Text {N=19 \; k=23} \\\$$
Wolfram Alpha: scripted link
Script
sum (binom(19,i)*(-1)^i*(19-i)^23) from i=0 to 18
This result equals the one presented in your book.
+89 Thank you for answer.
Where you find that formula my book not have it.
