schematic
twisted-iggy,
the b's indicate the connection is made to the base of the corresponding transistor (b T1 goes to the base of transistor T1 and so on). Connections are often drawn that way (using labeled nodes) to avoid cluttering up the schematic with wires.
You seem to have some confusion about the parts count. If you have high-efficiency LEDs (2 mA drive) a direct-drive approach works fine. Although CD4000 parts have limited output drive, it increases with increasing VCC. There's not enough output drive to run the circuit at 5V, but at 12V the outputs can supply ~ 3 - 8 mA, enough to drive high-efficiency LEDs. For this scenario, you'd need (1) 4017, (1/2) of a 4013, (8 ) diodes, (10) LEDs and (2) current-limiting resistors, in addition to the 555 section (output "0" doesn't need a diode, since it's not ORed with any other outputs). You should also include a 0.1uF bypass cap across the power pins of each IC.
ljcox's circuit requires more components, but will drive standard LEDs (10-20mA). My guess is that's what you have for LEDs, a simple test can confirm it.