The way to increase the LM3916's output current is to drive a PNP transistor with it:
That will work as a switch which switches the LEDs on and off. R1 prevents the transistor from conducting through the LM3916's leakage current. You can use a 10k\$\Omega\$ here.
If you want to control the brightness with your 0-15V you'll need to convert that voltage to a current; LEDs are current controlled devices.
Place the following circuit between the LEDs and ground:
The LED's current will be \$\dfrac{V_{IN}}{R_{SENSE}}\$
The TIP120 does not need 120mA at the base for normal operation, that's the absolute maximum rating, above which you don't want to go.
The spec you are mostly interested in is the hFE (current gain), which for a darlington is very high, since it's two transistors connected in a way so the current gains multiply.
For the TIP120 it's given as minimum 1,000 (compare with a typical 200 for a single bipolar transistor)
Also important are the max collector current (5A) and the collector emitter voltage (60V)
The main disadvantages are that the base-emitter voltage is doubled compared to a single transistor (~1.4V), and the saturation voltage is higher (typically ~0.8V compared to ~0.2V at low currents)
These points are rarely a problem for a simple switch driven from a micro pin. At higher collector-emitter currents though, the Vsat rises and can interfere with desired operation and cause problems with dissipation.
For example, in the TIP120 datasheet note that at 3A Ice, Vsat is given as 2V, but at 5A it has risen to 4V. That's 20W of dissipation, a lot to heat to try and get rid of to keep the temperature down. So when switching a large current you need to take these factors into account, and maybe decide to look at a more suitable part (e.g. logic level, low Rsdon power MOSFET)
Since we have a gain of 1000, we hardly have to draw anything from the micro pin. Let's say we want to switch 1 Amp:
1A / 1000 = 1mA into the base needed.
If we have a drive voltage of 5V, then we subtract the Vbe from the drive voltage and divide by the current:
(5V - 1.4V) / 1mA = 3.6k resistor. To give it a bit of leeway select something a bit smaller like 2.2k. This still only draws ~1.6mA.
I wouldn't read too much into the different prices - the price of components is often dictated by how popular they are, the more they sell the less they cost. If you see better specs at a cheaper price, go for it ;-)
You can some pretty odd prices when the component is scarce/new/obsolete - I saw a 10uF ceramic capacitor priced at £7.50 (qty 1) on Farnell the other week...
Best Answer
If you're selecting between two sources of binary digital data, you should be looking for a digital multiplexer. If you're selecting between two analog signals you should be looking for an analog multiplexer. There are dozens of part numbers of either type available from the major IC vendors, depending on the details of your requirements.
Buying either of these circuits as an IC will probably be lower cost (after accounting for assembly costs) and almost certainly be more reliable and give more consistent results than constructing them out of transistors.