Can the 74LS244 drive 8 LEDs or should I switch to ULN2803 ?

Hallo Christian,

if you have a look at datasheet of 74LS244, you will notice, that a low level current of less than 24mA per output is recommended. You will also find the specification about output voltage if output low current is 12mA and 24mA. Just have a look into datasheet. There's nothing myterious.
At IOL = 12mA we find VOLmax = 0.4V. At IOL = 24mA we find VOL = 0.5V.
To find out whether 74LS244 can sink your 8 LEDs via 330R resistors, we make the following estimation:

If voltage drop across LED is about 2V (which can vary between about 1.5V and 2.5V, by the way), then at 330R resistor about 3V is dropping. So, about 3V / 330R = 9mA is flowing through each LED. This current is well within the limits, what a 74LS244 can tolerate. So, no problems occur.
Even if voltage drop across LED would be unrealistically low, namely 1.5V, then no more than 10.6mA will flow per output, which is also well within the limits.

So, your circuit will work!

But, as you wrote, that you are eager to learn something new, let's join the following discussion.

Assume you use standard 5mm LEDs, like 590-446 (red) and 590-452 (green) from RS components. These are cheap standard LEDs, giving a brightness of 5mcd at 10mA. Voltage drop at this current is 2.0V for both. If you conmect them to 74LS244 outputs via 330R resistors, a current of I = U / R = (5V - 2.0V - 0.2V) / 330Ohm = 8.5mA will flow per output. 8 times 8.5mA plus the power supply current of 74LS244 of 46mA maximum gives you a total current consumption of 114mA! And as you want to sink 16 LEDs totally, this will give an overall current consumption of 228mA, only for the LEDs!! Do you really want this?

If you, on the other hand, choose high efficiency 5mm LEDs, like 585-573 (red) and 585-589 (green), which are only a bit more expensive, situation looks totally different. These LEDs give you a brightness of 30mcd (red) and 25mcd (green) at same current, namely 10mA, at voltage drop of 2.3V (red) and 2.3V (green).
The advantage is now, that these high efficiency LEDs only need 1/6 (red) and 1/5 (green) of the current of the above standard LEDs to produce the SAME brightness!! This helps you in dramatically decreasing total current consumption, without suffering from lower brightness.

Let's calculate the current limiting resistor needed for these high efficiency LEDs to produce the same brightness:

8.5mA / 5 = 1.7mA. And (5V - 2.1V - 0.1V) / 1.7mA = 1647Ohm. So, we can choose 1.6kOhm resistors.

Now, what about total current consumption?
If only 1.7mA are flowing per output, you need not to use 74LS244. A 74HC541, which I recommended you in my last reply, will also work. Big advantage: Static power supply current is totally negligible compared to 74LS244!!
So, the same circuitry but with high efficiency LEDs and 74HC541 will need only 27mA. This is only about 12% of that what's needed with standard LEDs and 74LS244!!!

And, is this a relevant improve???

Kai