Math CentralQuandaries & Queries


Question from Farihin, a student:

Lets say that i have keys, and each key is for notes of a musical instrument,
So i wanted to find out the number of notes i can get for a certain number keys,
of course in the form of an equation. Notes can use as many keys, it can use 1, or 2, or 3, or even 100.
Notes in real life is not as such, but ignore reality.
I tried doing this but i can't seem to find a formula for it.
For example, i have 4 keys, say A, B, C, and D.
so, for notes that uses one key are 4, which is A, B, C, and D themselves.
for notes that uses two keys are 6,
AB, AC, AD, BC, BD and CD.
for notes that uses three keys are 4,
lastly for notes that uses all four keys is 1, ABCD.
So, the total will be 4+6+4+1=15#

The nth term for the first equation is n, the second is [(n^2)-n]/2
the third and the fourth, i don't know but the final answer should be like,
n + [(n^2)-n]/2 + [3rd] + [4th]

Sorry for the long question though...

Hi Farihin,

I want to count the number of "notes" from the 4-key instrument you describe in a different way. Hopefully a way that shows you how to arrive at the general statement for an n-key instrument. As you did I am going to label the keys A, B, C, and D and for each of them you have the choice of leaving it closed or opening it. I am thinking of a wind instrument where each of the valves (keys) is closed unless you press on the valve to open it.

Start with the valve A. You can open or close it giving the two possible sounds, $A$ or $O.$ I am using $O$ to indicate that all the valves are closed and there is no sound. Now, regardless of the position of the valve A you can open or close valve B. Thus if A is open you can press B to get $AB$ or leave it closed to get only $A.$ If A is closed then you can press B to get $B$ of leave B closed to get $O$ again. Hence the possible notes using keys A and B are

$AB, A, B \mbox{ and } O.$

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

$ABC, AB, AC, A, BC,B, C \mbox{ and } O.$

Finally extend to a 4-key instrument using the D key. Each of the 8 possible notes listed above can be extended to the 4-key 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,


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