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# You have a bag of 9 letters: 3 Xs, 3 Ys and 3 Zs. You are

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You have a bag of 9 letters: 3 Xs, 3 Ys and 3 Zs. You are [#permalink]  11 Jul 2003, 18:43
You have a bag of 9 letters: 3 Xs, 3 Ys and 3 Zs. You are given a box divided into 3 rows and 3 columns for a total of 9 areas. How many different ways can you place one letter into each area such that there are no rows or columns with 2 or more of the same letter (one such way is shown below):

X Y Z
Y Z X
Z X Y

(A) 5
(B) 6
(C) 9
(D) 12
(E) 18
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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(B) 6, the logic being that you are essentially ordering 3 items and must account for all possible ways those items can be ordered
Director
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It also gussed it as 6. But, how to get the answer mathematically?
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Aha, I forgot to hold a mirror up to it. 12.
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JP wrote:
Aha, I forgot to hold a mirror up to it. 12.

You got the right answer, though, i don't think for the right reason. I'm not sure a "mirror" image will work. You need to rotate the second row one unit to get the second possibility for each first row possibiltiy. Holding a mirror will not change the letter under the "x" from a "y" to a "z". Here is a rather long winded explanation for something that is not really that complicated.

The correct answer is (D). This simplest way to solve this is solve this logically by taking one row at a time, and one square at a time in each row if needed. Let us start with the top row. In the top left area, we have the choice of 3 different letters. Next, in the top center area, we now have the choice of only 2 different letters so as not to match the first letter, and in the last box, we must use the last of the 3 choices of letters as not to match either of the letters in the first two areas. Hence, we have 3 x 2 x 1 or 6 ways to fill in the top row without duplicating a letter across it.

Now letтАЩs take a particular arrangement of the top row, say the one given in the example above. Since the top-left letter is an X, we could put either a Y or a Z into the 2nd row-left area. LetтАЩs put a Y in that space. Now, in order to maintain the constraint of not duplicating any letter in any row, we can only put an X or a Z in the center area. But if we put an X in the center area, we are forced to put a Z in the 2nd row-right area in order not to duplicate the previous letters in the 2nd row, but that would match the Z directly above, so this choice is not possible. However, if we put a Z in the center area, we are now forced to put an X in the 2nd row-right area and this would be consistent with our constraint and Y-Z-X is a possible combination for the 2nd row.

Now, letтАЩs say that we put a Z in the 2nd row-left area. Now we can only put an X in the center square so as not to match the Y above it or the Z to the left of it. This forces us to put a Y in the 2nd row-right area, which is still consistent with our constraints, and Z-X-Y is the only other possible combination for the 2nd row, i.e., there are exactly two possible 2nd rows: Y-Z-X and Z-X-Y. Similar logic can be performed for any of the 6 possible arrangement of the top row. This means that for each of the possible 6 arrangement of the top row, there are exactly 2 possible 2nd row arrangements, or 6 x 2 = 12 possible arrangement of the 1st and 2nd rows together.

Now letтАЩs examine the 3rd row given a particular 1st and 2nd row. Once again, letтАЩs use the specific example above. Note that if the 1st and 2nd rows are X-Y-Z and Y-Z-X, respectively, there is exactly one way to fill out the bottom row, namely Z-Y-X (just choose the letter that doesnтАЩt match the two above it). Since the 3rd row is completely determined by the 1st and 2nd rows, and there are only 12 ways to arrange both the 1st and 2nd rows, there are also only 12 ways to arrange all of the rows and the correct answer choice is (D).
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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To Akami [#permalink]  12 Jul 2003, 09:02
I've heard your a tutor, do you make up these questions on your own, or do you take it from a prep company?

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My "composed" questions are made up by me. Sometimes, they are variations off of published questions, but I usually add a "twist". These questions are purposely difficult (for GMAT questions) and are meant to give 95%ers and above a challenge compared to what I believe are easy problems in the prep booklets.

I also write sometimes long-winded explanations because I do not assume knowledge of advanced concepts or fancy formulas by my students. (Fancy formulas are just concise summations of basic concepts, but people who memorize formulas often forget the underlying concepts).
_________________

Best,

AkamaiBrah
Former Senior Instructor, Manhattan GMAT and VeritasPrep
Vice President, Midtown NYC Investment Bank, Structured Finance IT
MFE, Haas School of Business, UC Berkeley, Class of 2005
MBA, Anderson School of Management, UCLA, Class of 1993

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Re: composed problem #8 [#permalink]  10 Sep 2008, 06:58
I got 12 and here is my methodology.

A 1 2 3 B 1 2 3 C 1 2 3
x y z
Lets pick x,y,z for A. This set up is such that A will be the top row, B will go under it as the middle row and C will be the bottom row.

Now, x can go in 2,3
y can go in 1,3
z can go in 1,2

If you start plotting in B and C, you will notice that once x is plotted in 2 and 3 positions, there is only 1 way in which y and z can be plotted in B and C, the reason being that x and y have 1 common position in 3.
Lets plot now.

x, y, z z, x, y y, z, x

x, y, z y, z, x z, x, y

Therefore you find that each A gives 2 combinations of B and C. These 2 combinations, if you notice, have the same sequences but they are still considered 2 different combinations because we have fixed the positions of A, B AND C
Now, A can be formed in 3*2 = 6 ways and each way will give rise to 2 unique combinations. Therefore, the total no. of ways is 12.
Re: composed problem #8   [#permalink] 10 Sep 2008, 06:58
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