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

First 4 letters you only have Y's and Z's

so its \[\frac{ 8! }{ (4!)(4!) } \times \]

The next 4 you have no Y's so do the same thing and the same for the last 4 as well.

Note: the two 4 factorials represent the repetition of the Y's , Z's and X's.

Answer 2

so what do you get

Answer 3

so is it 8!/4!4! times 3?

Answer 4

what are your options really because you can't really make a word with only X Y and Z

yea...

Answer 5

not times 3 but times the same thing over three times like to the third power.

Answer 6

but it says a "word" with only X,Y,and Z's

Answer 7

lol I can't think of any word of the bat that starts with those letters haha.. anyhow what did u get?

Answer 8

no its just a "word" with any letters that you can get by using X, Y, and Z

Answer 9

But what about the conditions in the question above.... if conditions don't apply do this

12!/4!4!4! alright.

Answer 10

so the answer is 12!/4!4!4! right?

Answer 11

yes, if there are no conditions.