+0

# 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 le

0
1633
5

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?

Guest Mar 29, 2015

### Best Answer

#1
+85819
+10

I'll give this one a try  ....though it doesn't appear to be too easy!!!

Notice that we could have these 2 configurations right off the bat......

yyyy zzzz xxxx      and  zzzz xxxx yyyy

Next, we could have this

yyyz  xzzz xxxy  .......but....since each single letter within each grouping can appear  in one of four positions,

we have 4 x 4 x 4 = 64 possiblities

Next, we could have

yyzz xxzz xxyy ....notice that within each subgroup, we have (4!)/(2! * 2!)  = 6 recognizable "words"

So  6 x 6 x 6 = 216 possibilities

We could also have

yzzz xxxz xyyy...which is again, 64 possibilities

So...if I haven't missed any combos, the total possible "words" = 2 + 2(64) + 216 = 346

Could somebody else check this....(Melody, Alan, heureka ??)

CPhill  Mar 29, 2015
Sort:

### 5+0 Answers

#1
+85819
+10
Best Answer

I'll give this one a try  ....though it doesn't appear to be too easy!!!

Notice that we could have these 2 configurations right off the bat......

yyyy zzzz xxxx      and  zzzz xxxx yyyy

Next, we could have this

yyyz  xzzz xxxy  .......but....since each single letter within each grouping can appear  in one of four positions,

we have 4 x 4 x 4 = 64 possiblities

Next, we could have

yyzz xxzz xxyy ....notice that within each subgroup, we have (4!)/(2! * 2!)  = 6 recognizable "words"

So  6 x 6 x 6 = 216 possibilities

We could also have

yzzz xxxz xyyy...which is again, 64 possibilities

So...if I haven't missed any combos, the total possible "words" = 2 + 2(64) + 216 = 346

Could somebody else check this....(Melody, Alan, heureka ??)

CPhill  Mar 29, 2015
#2
+5

This is correct. I entered this answer. And I understand fully. Thanks CPhil!!

Guest Mar 29, 2015
#3
+85819
0

Thanks.....I get lucky every now and then on these "counting" problems..........LOL!!!

CPhill  Mar 29, 2015
#4
+92221
+4

Hi Chris,

Great work.  I haven't looked through your logic and i have not finished my own solution either.

I am going to do it here, and if I am really lucky my answer will be the same as yours.  :)

 FIRST 4 - UNSORTED POSSIBILITIES YYYY      1 WAY YYYZ      4!/3! = 4 WAYS YYZZ      4!/(2!2!) = 6 WAYS YZZZ      4!/(3!) = 4 WAYS ZZZZ      1 WAY LAST 8 POSSIBILITIES XXXXZZZZ     8!/(4!4!) = 70 XXXXZZZY     8!/(4!3!)=280 XXXXZZYY     8!/(4!2!2!)=420 XXXXZYYY      8!/(4!3!)=280 XXXXYYYY      8!/(4!4!)= 70 TOGETHER 1*70=70 4*280=1120 6*420=2520 4*280=1120 1*70=70

Total

\$\${\mathtt{70}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,120}}{\mathtt{\,\small\textbf+\,}}{\mathtt{2\,520}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1\,120}}{\mathtt{\,\small\textbf+\,}}{\mathtt{70}} = {\mathtt{4\,900}}\$\$

Oh dear - this doesn't look good.     Help!    Have I done something wrong?

Melody  Mar 30, 2015
#5
+1

We solve by casework based on where the X's are placed.
Case 1: All 4 X's in the last 4 letters. There is only 1 word possible (YYYY ZZZZ XXXX)

Case 2: 3 X's in the last 4 letters and 1 X in the second 4 letters. Then any word we build will be a permutation of YYYZ XZZZ XXXY, where we permute only within the blocks of four letters. There are thus 4^3= 64 possible words in this case.

Case 3: 2 X's in the last 4 letters and 2 X's in the second 4 letters. Then our possible words are permutations of YYZZ XXZZ XXYY, where as before we permute only within the blocks of four letters. There are thus

\({4\choose 2}^3 = 216\)

possible words.

Case 4: 1 X in the last 4 letters and 3 X's in the second 4 letters. By symmetry with Case 2, there are 64 possibilities.

Case 5: All 4 X's in the second 4 letters. By symmetry with Case 1, there is only 1 word possible (ZZZZ XXXX YYYY).

This gives a total of 1 + 64 + 216 + 64 + 1 = 346 words.

Guest Jul 1, 2016

### 8 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details