Yes, you can. Since she didn't specify what to do with invalid input, you can do what you wish. However using don't cares will result in a smaller result.
Do I combine all of the expressions (as in W+X+Y+Z or WXYZ)
No. What the equations show you are how to derive outputs for four separate pins. If you wanted to combine the four pins you should have done it earlier when coming up with the Karnaugh map (the Karnaugh map would then only have one output instead of four).
First step, treat the four as four separate circuits:
A ─┬───────────────┐
│ │
B ──┬─[NOT]────────[AND]┐
││ │ │
C ────[NOT]──┬─────┘ [OR]── W
││ └[OR]─┐ │
D ────────────┘ │ │
│└──────────────[AND]┘
└──[NOT]────────┘
.... implementation of other circuits left as homework
A ──[NOT]────────┐
│
B ──[NOT]──┐ [OR]─┐
[AND]─┘ │
C ──[NOT]──┘ [AND]── Z
│
D ────────────────────┘
Second step, join A to A, B to B etc.. This is obvious because obviously A is the same as A etc.
A ─┬────┬───────────────┐
│ │ │
B ──┬────┬─[NOT]────────[AND]┐
││ ││ │ │
C ───┬─────[NOT]──┬─────┘ [OR]── W
│││ ││ └[OR]─┐ │
D ────┬────────────┘ │ │
││││ │└──────────────[AND]┘
││││ └──[NOT]────────┘
││││
└───────[NOT]────────┐
│││ │
└──────[NOT]──┐ [OR]─┐
││ [AND]─┘ │
└─────[NOT]──┘ [AND]─ Z
│ │
└──────────────────────┘
.... implementation of other circuits left as homework
The circuit should be working at this point but if you're paying attention you may notice that some parts of the W and Z circuits are sharing the same logic. For example, we're using NOT on B and C twice. The next (optional) step is to refactor the circuit and remove redundant/repeated subcircuits and components:
A ───┬──────────────────┐
│ │
B ────┬─[NOT]──┬────────[AND]┐
││ │ │ │
C ──────[NOT]───┬──┬────┘ [OR]── W
││ ││ │ │
││ ││ [OR]─┐ │
D ─┬───────────────┘ │ │
│ ││ ││ │ │
│ │└─────────────────[AND]┘
│ │ ││ │
│ └──[NOT]─┬─────────┘
│ │││
│ └─────────┐
│ │└─┐ [OR]─┐
│ │ [AND]─┘ │
│ └──┘ [AND]─ Z
└─────────────────────────┘
.... implementation of other circuits left as homework
Best Answer
Write down your 10 inputs, D0 to D9 and the Excess03 code (4 outputs).
Table will have 10 lines. You will have 4 inputs (BCD) feeding 4 outputs.
The key here is the 4 OR gates can have as many inputs as required. Look at the 1's in output and can you see how to get outputs from inputs.