I'm trying to build a 12-letter word with 4 x's, 4 y's, and 4 z's. in how many ways can i build a word if there are no x's in the first 4 letters, no y's in the next 4 letters, and no z's in the last 4 letters?

We could answer this step by step:

Step 1: All X's in the last four letters. (, , ): (4!/4!)³ = 1 combination

Step 2: 3X's in the last 4 letters and 1 X in the second 4 letters. Then any word we make will be a permutation of YYYZ XZZZ XXXY, where we permute only within the blocks of four letters. There are (4!/3!)³ = 64 possible combinations.

Step 3: 2X's in the last 4 letters and 2 X's in the second 4 letters. Then the possible combination would be YYZZ XXZZ XXYY, where we permute only within the blocks of four letters.
There are (4!/2!2!)³ = 216 combinations

Step 4: 1 X in the last 4 letters and 3 X's in the second 4 letters. Just like in the Step 2, there will be 64 possible combinations. (YZZZ XXXZ YYYX): (4!/3!)³ = 64

Step 5: All 4 X's in the second 4 letters. Just like in Step 1, there is only one possible combination which is ( ): (4!/4!)³ = 1

Totaling each step will give us: 1 + 64 + 216 + 64 + 1 = 346 words.

Im sorry i cant answer it : ( im so sorry