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combinatrics problem...I think

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combinatrics problem...I think

by mikebarmy » Thu Aug 04, 2011 8:53 pm
Each stock is designed with a one, two, OR three letter code (of the 26 letter alphabet). The letters may be repeated and a different order constitutes a different code. How many different stocks is possible to uniquely designate with these codes?

Can someone please explain how to arrive at the answer...and also how to subtract codes that are the same (doesn't just doing a permutation computation allow ABA to be counted twices - with the two A's?). Thanks!
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by gmatboost » Thu Aug 04, 2011 9:38 pm
This is really a Multiplication Principle of Counting question. We're not really choosing or ordering anything since all the letters can be repeated.

1-Letter code: 26 options
2-Letter code: 26 choices for 1st letter, 26 choices for 2nd letter, 26*26 = 676 options
3-Letter code: 26 choices for 1st letter, 26 choices for 2nd letter, 26 choices for 3rd letter, 26*26*26 = ? options

Note that the GMAT never requires math as big as 26^3 to be calculated by hand, so this is not a realistic GMAT question.

This method does not double count any entries. ABA is just one of the 3-letter codes. It is the one and only one in which we chose A, then B, then A.

According to the question BAA and AAB are different codes than ABA, so we don't need to worry about double counting there.
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by bubbliiiiiiii » Tue Aug 09, 2011 9:19 pm
Note that the GMAT never requires math as big as 26^3 to be calculated by hand, so this is not a realistic GMAT question.
Hi GMATBOOST,

This is one of the problems from GMAT Prep. Please find the screen shot attached herewith.
1-Letter code: 26 options - Agree
2-Letter code: 26 choices for 1st letter, 26 choices for 2nd letter, 26*26 = 676 options
[Aren't we duplicating the combination AA. I believe the duplicate entries should be deducted from 26*26]
3-Letter code: 26 choices for 1st letter, 26 choices for 2nd letter, 26 choices for 3rd letter,
26*26*26 = ? options
[Same as for point 2, aren't we duplicating entries for AAA, BBB, ... Thus, I feel here also we need to deduct the duplicate entries]
Could you please help me understand where am I going wrong?
Attachments
Combinatronics.png
Regards,

Pranay
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by Frankenstein » Tue Aug 09, 2011 9:31 pm
Hi,
What do you mean by duplication? Could you elaborate because I don't see any mistake in Greg's post.
Btw, the options are given in such a way that we don't need to calculate 26^3. The answer is a sum of three number ending with 6. So, the answer will be ending with 8.(6+6+6 = 18)
There is only one such option.
Cheers!

Things are not what they appear to be... nor are they otherwise
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by gmatboost » Tue Aug 09, 2011 9:33 pm
Fair enough, I was wrong. I still find it very unusual that it would require 26*26*26.

Anyway, to answer your actual questions:

Let's just focus on the 2-letter code.

We have 26 choices for the first letter. A, B, C, D, E ...
We have 26 choices for the second letter. A, B, C, D, E ...

Imagine drawing (or perhaps you might want to actually do this) a 26x26 grid on a piece of paper.
To the left of the grid, write the letters from top to bottom, with one letter for each row.
On top of the grid, write the letters from left to right, with one letter for each column.

Create codes in each of the cells in the grid. The first letter will come from the letters along the lefthand side of the grid. The second letter will come from the letters along the top of the grid.

Now, this grid has 676 cells, representing each of the 2-letter codes.
Only one of these cells is AA.
There is a whole row of A_ codes, like AB AC AD, etc.
There is a whole column of _A codes, like BA CA DA, etc.
But there is only one code at the intersection of A and A: AA.

On a smaller scale, if you do this with three letters:

AA AB AC
BA BB BC
CA CB CC

3*3 = 9 codes, none of which are repeats.

When you expand to three letters, the situation is pretty much the same. There is only ONE way to get AAA, which is to select A for each of the three letters. There is no double or triple counting of AAA, because there is only one SET of actions that creates that code. Specifically: choosing A for the first spot, choosing A for the second spot, and choosing A for the third spot.

Let me know if this explanation is any better. Definitely do let me know if this isn't convincing.
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by bubbliiiiiiii » Tue Aug 09, 2011 9:39 pm
Hi Greg,

The problem now seems to be clear to me.

Thanks for the explanation that you have provided.

I think I overthought on this and made this one complex. Anyways, once again thanks for providing your explanation. Can you also guide me how to improve in tackling such questions?
Regards,

Pranay
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by gmatboost » Tue Aug 09, 2011 10:12 pm
The best tip I can give for questions with huge numbers involved is:

Try to answer the question with much smaller numbers, and then see how that applies to bigger ones.

In this case, you might answer the question as if A and B are your only choices.
A
B
AA
AB
BA
BB
AAA
AAB
ABA
ABB
BAA
BBA
BAB
BBB

And from there see what the rule is, and apply it to the larger number.
Greg Michnikov, Founder of GMAT Boost

GMAT Boost offers 250+ challenging GMAT Math practice questions, each with a thorough video explanation, and 100+ GMAT Math video tips, each 90 seconds or less.
It's a total of 20+ hours of expert instruction for an introductory price of just $10.
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