Hi Scott!! Your lessons have really changed my bass playing. Thank you for sharing all this top level stuff for free here on TH-cam. As soon as I start working I hope to join the academy because it really is incredible. I just wanted to throw this idea out there because I noticed you have a lot of subscribers but I think that there is not enough for how good your stuff actually is and that may be due to a language barrier. For example: I'm originally from Brazil, so I only started watching your videos after I moved to America and learned English. I just think what you are doing is so good that you should have millions of subscribers. Maybe (but I am not sure about this) having translations on a couple of main languages could double if not triple the number of subscriptions! God bless, man!!
Mathematically the answer is 4! x 12 keys (ignoring the double naming of some keys) = 288. However, as with triads, standard music theory only recognises the Root inversion (e.g. Cmaj7/C - root note in bass); 1st inversion (e.g. Cmaj7/E - major 3rd in bass); 2nd inversion (e.g. Cmaj7/G - perfect 5th in bass); 3rd inversion (e.g. Cmaj7/B - major 7th in bass) the order of the notes stacked above the bass note is ignored. So "standard" chord theory only give 4 inversions of a major 7th chord in 12 keys, so 48 in total. In other words there are two answers, either 288 (maths based answer) or 48 (standard chord construction theory answer).
Zon Brookes you totally beat me to this. When they were mentioning permutations - my brain went straight to factorial. 4! * 12 keys * 4 inversions. 24 * 12 * 4 = 288 * 4 = 1152
using the four notes in any order, there are 288 combinations possible. Using the standard inversions, the number of permutations in all 12 keys is 48.
well, if we don't count the root in an octave higher and stay within one octave then my answer is 288 possibilities. 6 permutations per inversion and no permutation is repeated. if we should use elements in octave positions the numbers multiply exponentially depending on how many elements we use within the range of multiple octaves.
There are six permutations for the arpeggio without inversiones (1357, 1375, 1537, 1573, 1735, 1753), then 3 inversions, so 6+6+6+6=24 permutations in just one key. For 12 keys, 12 times 24 equals 288.
permutations of a 7th chord in a single position, 4 notes available, is 4 * 3 * 2 * 1 (aka 4!) is 24, there are actually 15 -written- major keys, you mention 12 actual major keys. so, it's 12 * 24, and the way I can keep it in my head, is if I turn it into 12 * 2 * 12, and I know 12 * 12 is 144, times 2 is 288
There are 1152 permutations. I thought it was 288, but that's always playing the lowest note first. Taking root position as an example, we can play any of the four notes first. So it becomes 4x3x2x1=24. 24 permutations per position, 96 per tone and 1152 total.
There are 6 possible permutations, and 4 possible inversions. Times these two together = 24 (and that's just one chord!) Translate this to all 12 keys and you have 288. Wow, time to get in the shed! ;)
great video mate and now to the main thing I've calculated all the possibility of all maj7 in a single key which is 12 and multiplied it by 12(total no. of keys) which gives us a total of 144.
It's a shame that two different names are given for what the subject is here, namely 'chord tones' and 'arpeggios'. Beginners are likely to take some time to realise that they are one and the same thing.
To different parts of the same thing. The actual chosen notes that make a chord , and when they are played. Chord tones describes what tones are played. The arpeggio tells us when they are played.
Phil's from London, so it's a version of the classic "cockney" accent. It's still English, but not "The Queen's BBC English" (otherwise known as "Received Pronunciation"). You have to have lived in Essex, London, or the South East of England to understand it, and having met Phil and had a chat with him, he's a Diamond Geezer....
Hi Scott!! Your lessons have really changed my bass playing. Thank you for sharing all this top level stuff for free here on TH-cam. As soon as I start working I hope to join the academy because it really is incredible. I just wanted to throw this idea out there because I noticed you have a lot of subscribers but I think that there is not enough for how good your stuff actually is and that may be due to a language barrier. For example: I'm originally from Brazil, so I only started watching your videos after I moved to America and learned English. I just think what you are doing is so good that you should have millions of subscribers. Maybe (but I am not sure about this) having translations on a couple of main languages could double if not triple the number of subscriptions! God bless, man!!
Mathematically the answer is 4! x 12 keys (ignoring the double naming of some keys) = 288. However, as with triads, standard music theory only recognises the Root inversion (e.g. Cmaj7/C - root note in bass); 1st inversion (e.g. Cmaj7/E - major 3rd in bass); 2nd inversion (e.g. Cmaj7/G - perfect 5th in bass); 3rd inversion (e.g. Cmaj7/B - major 7th in bass) the order of the notes stacked above the bass note is ignored. So "standard" chord theory only give 4 inversions of a major 7th chord in 12 keys, so 48 in total. In other words there are two answers, either 288 (maths based answer) or 48 (standard chord construction theory answer).
Zon Brookes you totally beat me to this. When they were mentioning permutations - my brain went straight to factorial. 4! * 12 keys * 4 inversions. 24 * 12 * 4 = 288 * 4 = 1152
I learnt loads from Phil Mann at SBL. My favourite teacher there!
4* 3 *2 * 1 = 24 for permutations in a key x 12 keys = 288
Thanks Scott & Phil !! Great video ... very informative.
6 permutations per inversion. 4 inversions per 7th chord.
12 keys
That's (6×4)×12=X
24×12=288
using the four notes in any order, there are 288 combinations possible. Using the standard inversions, the number of permutations in all 12 keys is 48.
well, if we don't count the root in an octave higher and stay within one octave then my answer is 288 possibilities. 6 permutations per inversion and no permutation is repeated. if we should use elements in octave positions the numbers multiply exponentially depending on how many elements we use within the range of multiple octaves.
There are six permutations for the arpeggio without inversiones (1357, 1375, 1537, 1573, 1735, 1753), then 3 inversions, so 6+6+6+6=24 permutations in just one key. For 12 keys, 12 times 24 equals 288.
Exactly what I need to know more about, taking it to the next level. By the way, that bass....sweeeet!
4 x 3 x 2 x 1 (number of ways to order notes) x 12 (number of keys) = 288 permutations
Great video guys! Really enjoyed it! The answer is 288!
(4 notes x 3 permutations x 2 inversions) x 12 keys = 288 :)
Loved it! And the answer, as everyone else said, 288,
Years later this is so interesting still
Isn't it 4x3x2x1 for permutations of one position of a maj7, multiplied by 4 for all the inversions, multiplied by 12 for all the keys. So, 1152?
4x3 x2 x1 = 24 for permutations in a key x 12 keys = 288
permutations of a 7th chord in a single position, 4 notes available, is 4 * 3 * 2 * 1 (aka 4!) is 24, there are actually 15 -written- major keys, you mention 12 actual major keys. so, it's 12 * 24, and the way I can keep it in my head, is if I turn it into 12 * 2 * 12, and I know 12 * 12 is 144, times 2 is 288
Great stuff!
you can do 48 different permutations in all 12 keys if you don't allow repetition of the permutations during the inversions
There are 1152 permutations. I thought it was 288, but that's always playing the lowest note first. Taking root position as an example, we can play any of the four notes first. So it becomes 4x3x2x1=24. 24 permutations per position, 96 per tone and 1152 total.
Henrique Vieira I didn't realize this was an older video 😮
288 in Spain!
There are 6 possible permutations, and 4 possible inversions. Times these two together = 24 (and that's just one chord!)
Translate this to all 12 keys and you have 288. Wow, time to get in the shed! ;)
1152 ( 4 notes have 24 possible orders, for 4 inversions for 12 keys )
I can't come up with any. Perhaps this book would help me? ;)
288! a year of hard work for sure!
Any chance we can have Phil give us a lesson in resolutions?
nice bass! what type of fender is it?
I'm going to take a chance with 288. Cheers!
3072. Great chat with 2 great players. Nice bass BTW.
John Bettencourt hi, how did you come up with that number?
I figured 4 cord tones to the power of 4 multiplied by 12 keys. Probably wrong but I figured I would give it a try.
Correct it ;) you still have a shot ;) you can delete this whole comment and do your math ;) if you want to, of course...good luck
Szamtarto Zsolt - Attila maybe I'll get points for creativity.
I can't find this book anywhere
great video mate and now to the main thing I've calculated all the possibility of all maj7 in a single key which is 12 and multiplied it by 12(total no. of keys) which gives us a total of 144.
288, and now the trick is to practice them all in one day!!
288 of course
It's a shame that two different names are given for what the subject is here, namely 'chord tones' and 'arpeggios'. Beginners are likely to take some time to realise that they are one and the same thing.
arthurmee I can vouch for that lol. When I was just beginning, it took me a while to realize there wasn't much of a difference between the two.
To different parts of the same thing. The actual chosen notes that make a chord , and when they are played. Chord tones describes what tones are played. The arpeggio tells us when they are played.
Well not really, arpeggios are made of chord tones, they're not exactly the same thing.
288! Love the videos Scott, Im 18 and at uni doing music Production, but getting madly into Bass, and want to become the best :)
(4x3x2x1) x 12 = 288 ouch!
288 (4 factorial = 28 X 12 = 288)
1152 or 288
I would say 1152
Personally... 1 😂 I think I need this book
there are actually 30 keys, major minor and sharps and flats.....Anthony Jackson teaches this...
1152 right?
288, so many to learn
Hey Scott, you will get 288 Permutations!
144 x 2 with inversion 288
Look at all of us breaking the 'Bass players are stupid' stereotype. 😀 Anyway, 4x3x2x1=24 and 12 keys, so 288.
Cheers!
24
288 !!! Hope I'll win
My face is melted but I'm gunna say 288
I thought there were 15 major keys, C, 7 sharp, 7 flat.
I'm gonna go all out and scream 288.
1x2x3x4 = 24
24x12 keys = 288
1152
288! And yes I used my calculator haha
Man I'm 3 years late to the 🥳. Oh well 288!
288 is my answer
Yes, it's 288.
288?
144
12 key is 288
the anwser is 288
That'll be... 4!*12=288 wow
288!!!
288.
Dammit, I never thought of checking other peoples' answers. I'm a dumbass. ;) Anyway my personal HAL-9000 says 288.
288
288 ... I think. unless it's a trick!
Jonathan Griffin don't forget inversions. 24 Permutations for note order ( in time ), and 4 inversions ( in pitch ) and 12 keys. - 288 * 4 = 1152
My music teacher asked us this last semester
288
3072 daaaaamn
288. Is the answer....
No idea
2500+ permutations
I'm gonna say the answer is 288
The Answer to The question Is 288. See you scott. peace.
288 boom
could really use that book!
4! x 12 = 288
dude, 3, 3 permutations. so easy.
Too many for my slow brain to remember! lol (naw, I think its 1152)
1440 because there are technically 15 keys.
288
Greetings !
288 to easy fam
but i dont have a bass so i technically wont need the book yet😣
Yeah, but... jazz...
I find it really hard to listen to Phil, this distracts from the potential great info he has to share.
Nam Trab he speaks a divergent dialect of English than Scott
I mean different
He speaks with a regional accent and that has nothing to do with his intelligence, all you have displayed is your ignorance.
Phil's from London, so it's a version of the classic "cockney" accent. It's still English, but not "The Queen's BBC English" (otherwise known as "Received Pronunciation"). You have to have lived in Essex, London, or the South East of England to understand it, and having met Phil and had a chat with him, he's a Diamond Geezer....
288 :)
1152
288?
144
288
288 :)
1152
288?
144
288
144
288
144
288
144
288