TITLE: Best way to divide multiple equal groups into known unequal groups. Christmas Lights
QUESTION [9 upvotes]: I'm not a math guru, but I do like Christmas lights. 
I am helping set up a decent sized lighting display for my work, I have measured all outlines and props to figure out how many lights we are going to need for each area.
The lights come with 100 lights/string however, we would like to cut the strings into the appropriate number of bulbs per outline/prop and put connectors on the ends so we don't have a bunch of extra bulbs at the end of props (some only need 16 bulbs on a prop and we don't want to be limited in having to daisy chain everything using lit up bulbs in the strings. We want to splice in plain wire between props.)
I realize there are other factors such as voltage drop, amperage, etc. we need to think about (I'm getting help from an electrical engineer friend for that.)
But I am trying to figure out the best way to divide the 100 bulb strings without having more cuts than I need.
I was first thinking of making the largest cuts first (say I need 56) I now have 44 bulbs on that string, but I don't think it's the most efficient to then say I need a sting of 42 so then cut 2 off when I may have needed 2 sets of 22 that would have equaled 44 and used up the remaining bulbs on that string.
Without including the full strings of 100 that I need, here are the quantities and number of bulbs per string I need... what is the best way for me to start trying to divide these up while avoiding unnecessary cuts and splices?
$$\begin{array}{rl}
    \text{Quantity} & \text{Bulbs Per String} \\
    8     &      10\\
    1     &      16\\
    4     &      18\\
    1     &      20\\
    4     &      22\\
    1     &      23\\
    1     &      30\\
    4     &      34\\
    34     &      36\\
    1     &      38\\
    3     &      40\\
    2     &      42\\
    1     &      44\\
    1     &      47\\
    1     &      50\\
    2     &      51\\
    3     &      52\\
    4     &      54\\
    2     &      56\\
    1     &      58\\
    1     &      60\\
    1     &      76\\
    4     &      77\\
    1     &      78\\
    2     &      79\\
    3     &      80\\
    4     &      81\\
    2     &      96\\
    8     &      99\\
\end{array}$$

REPLY [4 votes]: So you need a total of $4,944$ lamps, which means at least $50$ strings.
It would be a hard task also for a computer to proceed "brute-force" and check
all possible combinations.
So I can suggest you just to proceed by "wise judgement" in a sort of a
"greedy" algorithm. You are not going to achieve a fully optimal result, 
but the bulk discount you will get should compensate for that.
a) Clearly the 96 and 99 pieces will go by themselves: that makes 10 strings.
b) Next, the 36 lamps sections being the largest majority (34 out of 105)
are those "leading the play". I would try and accomodate them  first.
The complement of 36 is 64. Two pieces of 36 leave a remainder of 28.
There are not many ways to reuse such remainders.
Therefore proceed and accomodate these two measures first of all, 
then pass to the remaining ones.
-- Addendum --
I have been pondering upon the fact that the problem would be optimally solved
by an "analogic gravitational computer": a  prismatic box of width $100$
filled with rods (e.g. pencils) of the given lengths. 

Once well shaken a number of times it will reach a (near) optimal filling.
And if the apex angle is progressively reduced to get a parallelepiped
 $100 \times 1 \times H$, and the material of the box is transparent
you can read the solution.
I deem that such analogic calculator will perform the greedy algorithm said above (more or less).