Question

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?

Answer 1

how do you do this question

Answer 2

please help

Answer 3

so there is 8 possibilities for each letter in the word

Answer 4

really?

Answer 5

but how is there 2 possibilities for it

Answer 6

since there is only 3 letters

Answer 7

and one you can't use

Answer 8

i mean there are 8 places for an x to be and there are 8 places for a y to be and there are 8 places for a z to be

Answer 9

ok

Answer 10

then what do you do

Answer 11

im not completely sure but i think you multiply the anount of possibilities together 8x8x8=512

Answer 12

so the answer is 512 possibilities?

Answer 13

i believe so

Answer 14

ok

Answer 15

i tried it but its wrong