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With only 1 valve you got 2 possible notes, $A$ and $O$ and each of them can be extended in 2 ways, using the B valve, giving $2 \times 2 = 2^2 = 4$ possible notes using 2 keys. Now include valve C. Each of the 4 possible notes you have using 2 valves can be extended to a 3 key instrument, using the C key, in 2 ways giving $2 \times 2 \times 2 = 2^3 = 8$ possible notes. They are
Finally extend to a 4key instrument using the D key. Each of the 8 possible notes listed above can be extended to the 4key instrument in 2 ways, either you press the D valve or you don't. Thus I get $2 \times 2 \times 2 \times 2 = 2^4 = 16$ possible notes. In more conventional mathematical language there are $2^4$ subsets of a set with 4 elements. You got 15 notes and I got 16 because I included the note that produces no sound. Using the music interpretation of this process, $O$ is a rest. I hope this helps, Penny  


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