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Arranging letters

by wydadi » Thu Aug 12, 2010 12:49 pm
Hi all,

I see this problem as an example in the Gmat Flashcards made by Eric of BTG.
I don't understand why I have always an answer different of the one given. So before telling Eric there is a mistake, I want to be sure it is.

Here is the question:
How many ways can the letters in TRUST be arranged?

There are no answers choices, however Eric's answer is 60. I don't find this result, do you?
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by Gurpinder » Thu Aug 12, 2010 12:52 pm
wydadi wrote:Hi all,

I see this problem as an example in the Gmat Flashcards made by Eric of BTG.
I don't understand why I have always an answer different of the one given. So before telling Eric there is a mistake, I want to be sure it is.

Here is the question:
How many ways can the letters in TRUST be arranged?

There are no answers choices, however Eric's answer is 60. I don't find this result, do you?
TRUST

5!/2! = 60.

for a problem like this one, always put the TOTAL / repeating group!
"Do not confuse motion and progress. A rocking horse keeps moving but does not make any progress."
- Alfred A. Montapert, Philosopher.
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by wydadi » Fri Aug 13, 2010 12:08 am
I don't get it yet.
Let's see how I answered:
in the word trust we have two similar letters T, which means every arrangement of the two T will count once since there is no difference between the Ts.
That's why I consider only one T and then I'll put the other one.
First, to rearrange the 4 letters TRUS, we have 4*3*2*1=24 possibilities.
Then to put the other T, we can put it in one of the following places :
-between T and R
-between R and U
-between U and S
-on the right of S

Finally the total is 24*4=96

What did I do wrong?
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by Gurpinder » Fri Aug 13, 2010 7:10 am
wydadi wrote:I don't get it yet.
Let's see how I answered:
in the word trust we have two similar letters T, which means every arrangement of the two T will count once since there is no difference between the Ts.
That's why I consider only one T and then I'll put the other one.
First, to rearrange the 4 letters TRUS, we have 4*3*2*1=24 possibilities.
Then to put the other T, we can put it in one of the following places :
-between T and R
-between R and U
-between U and S
-on the right of S

Finally the total is 24*4=96

What did I do wrong?
Ok I think your approaching the question with a wrong logic.

In the word TRUST, the TOTAL possibility of EVERY arrangement would be 5! - because there are a TOTAL of 5 letters.
Now, in 5! - we are counting the T twice, whereas the arrangement is suppose to be of EACH DISTINCT LETTER. So since the 2 T's are not distinct, you have to divide the 5! by 2!. Every other letter in the word is distinct.

Look at the word ATLANTA

3 A's, 2 T's,

So the answer to this one would be (do it before you look at the answer) : [spoiler]7!/3!2![/spoiler]

I hope this helps!!
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by selfmade » Fri Aug 13, 2010 8:48 am
Does this formula always work in arrangements problems ?
Total Items ! /Repetited Items !

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Aiming for 780
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by Gurpinder » Fri Aug 13, 2010 8:52 am
selfmade wrote:Does this formula always work in arrangements problems ?
Total Items ! /Repetited Items !

Thanks,
Arati
It is that simple ONLY for the arrangement of words and letters.

If you get into math combinatorics problems, then you have to know permutation and combination -- which apply the same principle in a slightly different way.
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by Brian@VeritasPrep » Fri Aug 13, 2010 3:05 pm
Hey wydadi,

Great question - I love when people take the time to try to think through alternate methods, and I think you can learn a lot by recognizing why they do or don't work.

One of my favorite strategies for GMAT questions is to test your theory, or try to find a pattern, using small numbers. Usually you'll be able to prove yourself right or wrong, and also be able to understand conceptually WHY you're right or wron,g which leads to much deeper understanding.

Since TRUST has quite a few possibilities (5! = 120, the answer is 5!/2! or 60, and you had 96...way too many possibilities to try to draw out), let's use something smaller - 4 letters with one repeat:

ABCA

We should be able to list all of those out by staying organized and trying new permutations one at a time:

ABCA
ABAC
ACAB
ACBA
AABC
AACB
BCAA
BACA
BAAC
CBAA
CABA
CAAB

There are 12 ways to do it, and we can't come up with any more without just flipping our As, which doesn't give us anything new. 12, also, is consistent with the formula (which it should be, of course): 4!/2! = 12.

Now let's look at your method:

ABC gives you 3! ways to arrange, and then you can put the A in different spots: left of the A, left of the B, left of the C, and right of the C (AABC, AABC, ABAC, ABCA). This should point out a problem, though - the first two are the exact same! So you can see where your method won't work - every few permutations you'll end up with a duplicate spot, so you'll overestimate the number of possibilities.
Brian Galvin
GMAT Instructor
Chief Academic Officer
Veritas Prep

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