Aren't you assuming that there is no drop across each LED ?

The drop across each LED is why there's two different values of resistor used.

Imagine the above circuit without LEDs 3 & 4.

There are then three possible paths (when the input state is used to disconnect certain pins) for current to take:

1. Pin1 to Pin2 through LED1, where there is a total of 317 ohms of resistance.

2. Pin 2 to Pin3 through LED2, where there is a total of 317 ohms of resistance.

3. Pin 1 to pin 3 through LEDs 1 & 2, where there is a total of 94 ohms of resistance.

Now go to the LED array wizard:

http://led.linear1.org/led.wizEnter 5v source, 2.1fv, 10mA, and 1 LED. It tells you you need a 330 ohm resistor.

Now enter the same, but for 2 LEDs. It tells you you need an 82 ohm resistor.

Both these cases will light the LEDs with 10mA. Both cases will be just as bright. But in the case of 2 leds, more power is used to generate light, and less power is wasted in the form of heat, in the resistor.

To get the values in the above circuit, I did 82/2 = 41 and chose the closest value resistor I had on hand, 47 ohms.

I then subtracted 47 ohms from 330 ohms, and got 283 ohms. The closest value to that being 270.