I see these LED calculators that tell me to connect a single LED to one resistor like for output 12 I have 10 LED's and each is connected to a 500-550 ohm resistor. I just wanted to know if there is a way of connecting all the led's "neg pos neg" and use just one resistor at the end or at least 5 LED's?
yup thats how its done.
However, USB has limited current. Someone should correct me, but its somewhere around 150mA.
Heres how it works: LED has a 2V drop, say. So 3 LEDs connected in series (neg pos neg as you said) will drop 6V. if they are the 20mA kind of LEDs, they will take 20mA and 6V for all three. You could then add another, identical string, in parallel with that one. then you will need 40mA and 6V. However, if you hooked up the two strings end-to-end (series, neg pos neg), then you would need 12V and only 20mA.
Power stays constant. V * A = P, so you can change your parallel or series combination to get the amount of current and voltage. i.e. USB has 5V and 150 mA (I think, double check that), so you need to find the ideal combination of parallel and series strings to maximize that.
Additionally, LEDs rated at 20mA will probably still look about the same brightness at 15mA, so you might run them a little bit under spec to save power/get more LEDs. (simply use a slightly larger resistor on each string. change the current tolerance in those calculators you use to figure it out)