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?
@satellite73
so i need to ask satellite73?
Well he is the Legend.
but you can only send him messages if your his fan
i'm his fan but i can't send messages too him
i have no idea how to do this, but we could at least try something
well, here is the Legend.
cool
but how
you have four slots in which to put only y and z
ok
for some reason i can't send you a message even though i'm your fan
just off the top of my head, how many ways can this be done? 1) all y: 1 way 2) 1 z 3 y: 4 ways 3) 2z 2y: 6 ways 4) 3z 1 y : 4 ways 5) all z : 1 way so far we have \(16=2^4\) possibilities
so its 16?
oh hell no, we have lots more work to do
can you fan me so i can send you questions
what about the next 4 slots ? it can have no y, only z and x so i guess we have to work on what happened above
ok
if 5) we used up all the z there is only 1 way (all x) 4) used up 3 z there are 1z, 3x : 4 ways 3) used up 2 z : 2z, 2x : 6 ways 2) used up 1 z : 3z, 1z, 4 ways 1) used up no z : 4z 1 way
there must be an easier way to do this....
i think so too
well maybe not a tree diagram might help
but how
Is this right: The first 4 letters will be y and z. There are 16 arrangements of those 4 letters. The next group of 4 will be X and Z and there will be 16 arrangements. The last group of 4 will be X and Y and there will be 16 arrangements. So??? 16^3
so the answer is 16^3?
but its wrong
i tried it already
yeah i think there is something else going on here
if you make a tree diagram of sorts you might see it, then add up the possibilities use pencil and paper
i tried that but i didn't get anything
Well then it might be that some of the arrangements look the same because the letters are repeated .
then what can you get
Did you try 2^12 since there are 2 choices for each position?
yeah and it didn't work
got it its 346 ways
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