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Mathematics 16 Online
OpenStudy (anonymous):

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?

OpenStudy (bahrom7893):

@satellite73

OpenStudy (anonymous):

so i need to ask satellite73?

OpenStudy (bahrom7893):

Well he is the Legend.

OpenStudy (anonymous):

but you can only send him messages if your his fan

OpenStudy (anonymous):

i'm his fan but i can't send messages too him

OpenStudy (anonymous):

i have no idea how to do this, but we could at least try something

OpenStudy (bahrom7893):

well, here is the Legend.

OpenStudy (anonymous):

cool

OpenStudy (anonymous):

but how

OpenStudy (anonymous):

you have four slots in which to put only y and z

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

for some reason i can't send you a message even though i'm your fan

OpenStudy (anonymous):

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

OpenStudy (anonymous):

so its 16?

OpenStudy (anonymous):

oh hell no, we have lots more work to do

OpenStudy (anonymous):

can you fan me so i can send you questions

OpenStudy (anonymous):

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

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

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

OpenStudy (anonymous):

there must be an easier way to do this....

OpenStudy (anonymous):

i think so too

OpenStudy (anonymous):

well maybe not a tree diagram might help

OpenStudy (anonymous):

but how

OpenStudy (mertsj):

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

OpenStudy (anonymous):

so the answer is 16^3?

OpenStudy (anonymous):

but its wrong

OpenStudy (anonymous):

i tried it already

OpenStudy (anonymous):

yeah i think there is something else going on here

OpenStudy (anonymous):

if you make a tree diagram of sorts you might see it, then add up the possibilities use pencil and paper

OpenStudy (anonymous):

i tried that but i didn't get anything

OpenStudy (mertsj):

Well then it might be that some of the arrangements look the same because the letters are repeated .

OpenStudy (anonymous):

then what can you get

OpenStudy (mertsj):

Did you try 2^12 since there are 2 choices for each position?

OpenStudy (anonymous):

yeah and it didn't work

OpenStudy (anonymous):

got it its 346 ways

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