Suppose that I have a very special Ukulele that contains N frets. I can use atmost 4 fingers to play a chord. I am required to determine the total number of different chords that I can play.

Mattie Monroe

Mattie Monroe

Answered question

2022-10-27

Total Number of different playable chords
Suppose that I have a very special Ukulele that contains N frets. I can use atmost 4 fingers to play a chord. I am required to determine the total number of different chords that I can play.
Note: Two chords are different if one of the frets that are pressed in one of the chords is not pressed in another chord.

Answer & Explanation

Bobby Mcconnell

Bobby Mcconnell

Beginner2022-10-28Added 8 answers

Step 1
The first interpretation is that all the strings are tuned the same and so having a fret pressed on one string is considered the same as if it were pressed on the other string. The question reworded in more math-heavy terms becomes "How many subsets of size 4 or less exist for an N element set?"
Well... there are ( N k ) subsets of size k of an N element set, so our answer here is simply ( N 0 ) + ( N 1 ) + ( N 2 ) + ( N 3 ) + ( N 4 ) , the number of ways of pressing zero frets or pressing one fret or pressing two frets, etc...
Step 2
The other interpretation that seems natural enough is that each string is tuned differently and so the chord that is made by pressing (or not pressing) frets is also dependent on which string it was that the fret was pressed on. Ukelele typically have four strings, one for each available finger. Here, we ask "Which fret is pressed for the first string? Which fret is pressed for the second string? etc..." keeping in mind that not pressing a fret on that string is also an option. The question reworded in more math heavy terms becomes "how many functions are there from a 4 element set to an N + 1 element set?"
As each finger has N + 1 options (each fret being an option and the additional option of not having pressed a fret), we apply the rule of product to arrive at a total of ( N + 1 ) × ( N + 1 ) × ( N + 1 ) × ( N + 1 ) = ( N + 1 ) 4

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