Combinatorics problem with connecting computers A system administrator has to connect 16 computers to 4 network switches. Exactly 3 computers must be connected to the first network switch. 


*

*Show that there must exist a switch that has at least 5 connected
computers.

*In how many ways can we connect the computers if we know that they
are all different?

*In how many ways can we connect the computers if we know that they
are all the same?


I tried solving the third question with first finding the $3$-combinations for all $16$ computers: $\binom{16+3-1}{16} = 153$ ways and then for the remaining $3$ switches I used the equation: $x_1+x_2+x_3=13$, with $x_1$ having at least $5$ connected computers, meaning $x_1>4$. The solution I got was $84$ ways. $153+84$ ways $= 237$ ways. 
I don't know if this is the right approach. I would like to know if I am on the right path and what am I doing wrong. 
 A: We are told that exactly three computers are connected to the first network switch.  That means $16 - 3 = 13$ computers are connected to the remaining three network switches.  If we let $x_i$ be the number of computers connected to the $i$th network switch, then 
$$x_2 + x_3 + x_4 = 13$$
which is an equation in the nonnegative integers.  A particular solution of the equation corresponds to the placement of $3 - 1 = 2$ addition signs in a row of $13$ ones.  For instance,
$$1 1 + 1 1 1 1 1 1 + 1 1 1 1 1$$
corresponds to the solution $x_2 = 2$, $x_3 = 6$, $x_4 = 5$.  The number of such solutions is 
$$\binom{13 + 3 - 1}{3 - 1} = \binom{15}{2} = 105$$
since we must choose which two of the $15$ positions required for $13$ ones and two addition signs will be filled with addition signs.  Thus, there are $105$ ways to connect $16$ identical computers to the four network switches if exactly three of those computers are connected to the first network switch.
It looks like you were misled by the wording of the first part of the question.  If exactly three computers are connected to the first switch, then $13$ computers must be connected to the remaining network switches.  At least one of those three switches must be connected to at least five computers, otherwise there would only be at most $3 + 4 + 4 + 4 = 15$ computers connected to the network switches, a contradiction.  Since exactly three computers must be connected to the first network switch, the case in which $x_1 > 4$ is ruled out.  
