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I need this answer because it applies to a music theory question I have.

Lets say there are 12 on-off switches in a circle. How many unique possibilities are there when isometric or similar combinations are considered the same and say exactly 7 switches must be turned on? Such as:

on, on, on, off, off, off, off, off, on, on, on, on

is the same as

off, off, off, off, off, on, on, on, on, on, on, on

wow I would be very impressed and thankful if anyone has this solution and could explain it!

2007-03-01 00:12:17 · 4 answers · asked by Steven 2 in Science & Mathematics Mathematics

4 answers

Okay, first lets ignor the circle.
Now, the number of possible combinations is 12!/(7!5!).
Now, since its in a circle, you will have repetitions due to rotations. In other words , you could have the same order rotated 12 times
So, since there would be 12 rotations, you divide by 12.
Now, you need to think about the repetions due to the direction.
In other words, you would have the same order in 2 different directions, so you divide by 2.
So, that gives you ((12!/(7!5!))*(1/12)*(1/2) = 33.
So, I think there is 33 different combinations.

2007-03-01 00:56:03 · answer #1 · answered by yljacktt 5 · 0 0

Interesting problem... Since this applies to music, it looks like you want to know how many possible chords can be formed out of the 12 tones, regardless of key. (In what follows the 12 bits refer to the 12 notes from C to B.)

For example, 100010010000 (C- major) and 001000100100 (D-major) would count as the same...

...but reverse-order combinations would not be considered the same, e.g. 100010010010 (C-7th) is not the same as it reverse, 010010010001 (F-minor 6th).

So... How many chords are there?

1) With zero notes there is 1 possibility (silence).

2) With one note there is also 1 possibility (single note)

3) With two notes (intervals) there are 6 possibilities, from the half-tone to the tritone (110000000000, 101000000000, 100100000000, 100010000000, 100001000000, 100000100000).

4) With 3 notes you start frmo the six two-note possibilities and find a place to add the third note:
110000000000 → you can add the extra note in positions 3, 4, 5, 6, 7, 8, 9, 10 and 11 (not 12 because that 's a repeat of 1110000000000) → 9 possibilities.
101000000000 → positions 5 to 10 are good (others will cause a repeat) → 6 chords
100100000000 → positions 7, 8 and 9 are good.
The rest will only cause repeats. Total = 18 three-note chords are possible.

That's as far as I can go now...

2007-03-01 01:44:51 · answer #2 · answered by Anonymous · 0 0

let us consider the spacing between 2 consequtive offs as variable which can vary from 0 - 7 and there are four such spacings. To refine the problem a bit more, let the spacings be s1>= s2>= s3>= s4 >= 0 such that
s1+s2+s3+s4<=7
The no of solutions to this equation will give the no. of disjoint isometrics. No symmetric solution is possible in this case. So just multiply the no. of solutions by 12. (isometrics)
To ease the solution Substitute s4=t1, s3=s4+t2 s2=s3+t3 s1=s2+t4 to remove one of the constraints.
4t1 +3t2+2t3 + t4<=7 where all t 's >= 0. I think there is a way to get the no of solutions I just dont remember I will be back.

2007-03-01 01:13:29 · answer #3 · answered by Keeper of Barad'dur 2 · 0 0

I answered your earlier question. Is 13 the wrong answer ?

2007-03-01 00:53:17 · answer #4 · answered by nayanmange 4 · 0 0

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