How many 8 letter words are there using the letters X, Y, Z containing at least one X and two Ys?

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Not sure how to approach this question. I have $3^8$ total possible words, but obviously some of these don't meet the requirements. So now I want to figure out how many of these don't meet the requirements and subtract that from the total number. If anyone could give some indication on how I can get this number I would greatly appreciate it.

METHOD FOR ANSWERING:

  1. Find the total number of possible words. Which is $3^8$
  2. Use the inclusion-exclusion principle with the 0X, 0Y and 1Y. These are all ways in which the words created won't satisfy the mentioned restrictions in the question. $|0X|$=$|0Y|$=$2^8$. Since we have 2 remaining choices for the 8 remaining slots. $|1Y|$ = $8*2^7$ Since there's 7 slots to be filled with Xs and Zs and the Y could be in any of the 8 slots so multiply by 8 to ensure all possibilities are counted. Then consider the intersection of 2 elements. The ones with 0Y and 1Y you don't have to calculate since it's an impossible scenario. However when we have 0X and 0Y we know there is only one possibility remaining when they're all Zs.
  3. Once you've calculated everything you can simply subtract the value you just calculated away from the total value leaving you with your answer.