This exercise will inspire you and open up many possibilities for your rhythm guitar playing. The full lesson is here!
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I believe the answer to the question of how many combinations there are of 8th note strumming patterns is 8! = 40,320.
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8 Factorial (8!) , for those wanting to get Mathies 
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Bjeezus… yeah maybe I won’t do a book with ALL the patterns then 
lol!
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This is all based around the assumption that your strumming pattern is only one measure long. Theoretically there are infinite possibilities if you don’t limit yourself to repeating one measure patterns
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I believe the answer is actually 255. We’re using quarter notes, not eighth notes, but you can either strum or not strum at each position.
D U D U D U D U
2x2x2x2x2x2x2x2 = 256 but you can’t make a beat with absolutely no sound, so 00000000 is not an option.
If you move to sixteenth notes, you have 2^16 -1 = 65,535
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255 is the correct answer. If you want to have some fun and practice random patterns you can use the calcultor function in windows. Switch to the programmer view and select a number > 128 (so you have a down strum on beat 1) and <= 256 then look at the binary conversion. Example 172 = 1010 1100 in binary this would be D_ D_ DU _ _. Strum the strings on the 1’s.
193 = 11 00 00 01 DU _ _ _ _ _ U.
Computer geeks (like me) would understand.
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I found this exercise hilariously frustrating 
Every time I wrote down a pattern, played it and thought “yeah I like that one” I would realize that it was either the Old Faithful pattern D-DU-UD- or the Distraction pattern DU-U-U-UDU. I did that around 4 times before I could actually come up with a pattern that wasn’t one of those.
I guess the past week of judiciously practicing those two patterns with distractions has paid off, that I’m writing and strumming them subconsciously even when I don’t want to
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A post was split to a new topic: Sharing a Python script for generating an 8th strumming pattern
Justin’s free Strumming Machine is an excellent tool for learning various patterns. 
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This was a very good lesson, as always.
A great addition would be a a list of easy songs that use particular strumming patterns, or a link to a list that has that info.
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I almost agree. It’s 2^8, not 8!, but there’s no reason not to have a completely silent bar.
If you limit yourself to strumming the first down-beat, it would ‘only’ be 128.
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The answer is actually 256 patterns. 8! is the number of possible ways to arrange a set of 8 numbers, but we’re not rearranging the beats, just turning them off or on.
This is more like flipping a coin 8 times in a row and strumming every time there is a heads and not strumming when it is tails. So the answer is actually 2^8 power which is 256. Although not strumming at all (if you flip all tails) isn’t really a pattern, so instead the answer is 255.
If you instead only count the patterns that have a strum on beat 1, then it is 2^7 or 128. You can definitely fit that in a book Justin!
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l can confirm the 255. Basically what we’re asking is how many eight-digit sequences there are that consist only of 0’s and 1’s", excluding 00000000. As an example, here’s Old Faithful: 10110110. And here’s just downstrumming on the beat: 10101010. To count how many sequences there are, note that you’ve got 8 slots and defining a strumming pattern amounts to choosing some of those slots to have 1’s. How many ways are there, for example, to choose just one slot to have a 1 (e.g. 10000000, 01000000, 00100000, etc.). The answer is 8. How many ways are there to choose 8 slots to have a 1. Just one way to do it: 11111111. For two slots it’s a little harder, but the answer is something called “8 choose 2” which is equal to 8!/(2!6!) = (8*7)/2 = 28. For three, it’s “8 choose 3”, and so on. To get the total, you add all of these up: (8 choose 1) + (8 choose 2) + (8 choose 3) + (8 choose 4) + (8 choose 5) + (8 choose 6) + (8 choose 7) + (8 choose 8) = 8 + 28 + 56 +70 + 56 + 28 + 8 + 1 = 255. That’s it!
You might notice the symmtry. For example, once we’ve calculated 8 choose 2, we get 8 choose 6 for free. That’s because choosing 6 slots to have a 1 is the same as choosing 2 slots to have a 0. Same for 8 choose 3 and 8 choose 5, and so on.
Another approach to this is to think of it as a binomial tree. Each time the tree branches, the number of nodes doubles. This leads to the expression 2^8-1 that some of the other commenters have used.