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GMATBarcelona CAT - Counting methods

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GMATBarcelona CAT - Counting methods

by molt_llest » Sun May 11, 2008 1:01 am
Thank you for your help, I don't know how to solve this problem.

How many different possible arrangements can be obtained from the letters G,M,A,T,I,I, and T, such that there is at least one character between both I's?

a.- 360
b.- 720
c.- 900
d.- 1.800
e.- 5.040

Correct answer: c (answer in white color, drag over to see the answer)
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by arorag » Sun May 11, 2008 3:29 am
My ans is d.
total no. arrangements= 7!/2
no. of arrangements in which two I are together is 6!

No. of arrangements in which I are sep. by atleast one character= 7!/2-6!= 1800
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by sandeep_chhabra » Sun May 11, 2008 3:37 am
According to me

The total No of arrangements Possible (Even without NO character between 2 'I's) = 7! /(2! 2!) = 1260

Now the total arrangements where 2 'I's are TOGETHER = 6!/2! = 360

So the total arrangements where 2 'I' are NOT TOGETHER = 1260-360 = 900 Answer C

Correct me if I am wrong.
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by Magellan » Sun May 11, 2008 3:51 am
sandeep_chhabra wrote:According to me

The total No of arrangements Possible (Even without NO character between 2 'I's) = 7! /(2! 2!) = 1260

Now the total arrangements where 2 'I's are TOGETHER = 6!/2! = 360

So the total arrangements where 2 'I' are NOT TOGETHER = 1260-360 = 900 Answer C

Correct me if I am wrong.
I think you are correct. I guess the 2!2! is because you have 2T and 2I.

Could you please explain what it the 'theoretical' formula to be used?

For example: If I have 4 letters A/B/C/C --> total number of arrangements if 4!/2!= 12 ?

Thanks
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by molt_llest » Sun May 11, 2008 8:06 am
sandeep_chhabra wrote:According to me

The total No of arrangements Possible (Even without NO character between 2 'I's) = 7! /(2! 2!) = 1260

Now the total arrangements where 2 'I's are TOGETHER = 6!/2! = 360

So the total arrangements where 2 'I' are NOT TOGETHER = 1260-360 = 900 Answer C

Correct me if I am wrong.
Thank you for your answer. Could you explain how could you get to 7!/(2!2!) (I think it is 2!2! because of the double I and double T but I'm not sure) and how could you get the 61/2! (I don't have any clue how you reached here)
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by sandeep_chhabra » Sun May 11, 2008 10:06 am
molt_llest wrote:
sandeep_chhabra wrote:According to me

The total No of arrangements Possible (Even without NO character between 2 'I's) = 7! /(2! 2!) = 1260

Now the total arrangements where 2 'I's are TOGETHER = 6!/2! = 360

So the total arrangements where 2 'I' are NOT TOGETHER = 1260-360 = 900 Answer C

Correct me if I am wrong.
Thank you for your answer. Could you explain how could you get to 7!/(2!2!) (I think it is 2!2! because of the double I and double T but I'm not sure) and how could you get the 61/2! (I don't have any clue how you reached here)
molt_llest you are absolutely right there. 7!/(2! 2!) - here 2! is for 2 'I's and another 2! is to 2 'T's.

for 6! - actually we considered the 2 'I' to be ONE ENTITY because we want them together and thus counted them as one so the remaining letters that we are left with are 6. Now we divide 6! by 2! because of double 'T's.

Hope i was able to explain it :)
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by molt_llest » Sun May 11, 2008 12:20 pm
sandeep_chhabra wrote:
molt_llest wrote:
sandeep_chhabra wrote:According to me

The total No of arrangements Possible (Even without NO character between 2 'I's) = 7! /(2! 2!) = 1260

Now the total arrangements where 2 'I's are TOGETHER = 6!/2! = 360

So the total arrangements where 2 'I' are NOT TOGETHER = 1260-360 = 900 Answer C

Correct me if I am wrong.
Thank you for your answer. Could you explain how could you get to 7!/(2!2!) (I think it is 2!2! because of the double I and double T but I'm not sure) and how could you get the 61/2! (I don't have any clue how you reached here)
molt_llest you are absolutely right there. 7!/(2! 2!) - here 2! is for 2 'I's and another 2! is to 2 'T's.

for 6! - actually we considered the 2 'I' to be ONE ENTITY because we want them together and thus counted them as one so the remaining letters that we are left with are 6. Now we divide 6! by 2! because of double 'T's.

Hope i was able to explain it :)
Perfect!!! thank you very much, now is all clear.
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by Stuart@KaplanGMAT » Sun May 11, 2008 6:56 pm
The generic forumla for word jumble questions is:

n!/r!s!t!...

in which n is the total number of letters and r, s, t, etc... are the number of duplicate letters.

For example:

How many different "words" can be made out of:

DESERT = 6!/2!

DESSERT = 7!/2!2!

DESSERTS = 8!/3!2!

DESSERTED = 9!/3!2!2! (hey, it's real word! "Last night after the main course, I got desserted by the waiter." :D )

Of course, this formula applies to all permutation questions in which you use all the entities and not all of them are distinct.

For example:

In how many different ways can 3 pennies, 2 dimes and 2 quarters be arranged, if coins of the same type are identical?

Answer: 7!/3!2!2!
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