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?
+128473 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 ??)
+128473 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 ??)
+128473 Thanks.....I get lucky every now and then on these "counting" problems..........LOL!!!
+118608 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?
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.
