I'm trying to count these notes in 4/4 but am having trouble with which note gets the number. It goes 8th note, quarter note, 8th note, quarter note, followed by 2 8th notes. I understand it fits the 4/4 measure but which note gets counted with the number? I understand the e & a counting method but do I start with 1 on the first 8th note? This confuses me because by the time I get to the second 8th and quarter note, the count is all off. I'll include a photo of the measure to better explain it! Thanks! enter image description here

  • 1
    If have some troubles to match the given score with the piano tag. I would have chosen something like rhythm.
    – guidot
    May 20, 2020 at 15:14

2 Answers 2


The first measure only has 8th notes as the fastest note. The bar is a measure of 4/4 so there are four downbeats and 8 possible 8th notes. This would be counted as :

1 & 2 & 3 & 4

The e & a counting works for 4 notes per downbeat (1 e & a ), so it works here only if you count

1 e & a 2 e & a

Which would be a measure of 2/4 with 16th notes.

Counting the first bar with the first method (four counts):

1 & 2 & 3 & 4 & ( ONE AND two AND THREE and FOUR AND)

With the second way of counting the first bar (two counts):

1 e & a 2 e & a (ONE E and A TWO e AND A)

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    Yes, I think I have a pretty good understand of counting but for a moment I forgot that you won't necessarily have to say each number in each example. Sorry for my lack of terminology, I'm fairly new to this and teaching myself. Thank you for the help tho!
    – matt
    May 20, 2020 at 14:55
  • Are you sure about the count for bar two? Don't think that counts triplets.
    – Tim
    May 20, 2020 at 14:56
  • @Tim I only answered regarding the first bar (which I think was the only one in the question). The second bar would need a triplet count and totally different yes (1 & a 2 & a ..).
    – hirschme
    May 20, 2020 at 14:58
  • I only asked because of the last two sentences. And 1&a2&a won't do it, at the same counting speed as 1e&a.
    – Tim
    May 20, 2020 at 15:00
  • @Tim I addressed the two ways of counting the first bar (8th note count or alternate 16th note count, which would be less recommended but addressed by the original question)
    – hirschme
    May 20, 2020 at 15:01

What I used to do (and still do on occasions) is to find the shortest note, and give that a count of one. Here, there is a quaver that can be one. Not the first beat of the bar, but a count of one. So now, the first bar can be counte up to 8.

Since the first note is one count, the next is actually two, it counts like this: 1 23 4 56 7 8. Bold is what gets played where.

That second bar with triplet crotchets (1/4 notes) is more tricky. The minim (1/2 note) is easy, with an 8 count it gets played on 5678. But the first part doesn't really play in time counting to 8. Problem is there would be a count of 1234, and you have to play three notes equally in that time!

It won't even work with 1e&a2e&a counting! Until you can feel how triplets sound, a safe way is to split just that bar into a count of 12. Yes, it sounds weird, and the count will not be the same as the first bar, but just to get the timing correct, count 12 34 56 78 9 10 11 12.

I'd have honestly thought you may have asked about that bar too!

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