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Why cancel out the 2 S's and C's? (Permutation/Combination)

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Why cancel out the 2 S's and C's? (Permutation/Combination) [#permalink] New post 09 Dec 2012, 00:23
Here is the question:
What fraction of seven lettered words formed using the letters of the words CLASSIC will have the two C's always together?
A) 2/7
B) 5/7
C) 15/19
D) 4/19
E) 2/8


The answer explanation given is:
This is a typical permutation-probability problem. To make this problem easily understandable we will break into two parts:
i. First, we will find out all the seven lettered words from the letters of word CLASSIC
ii. Next, we will find out how many of these words will have the two C's together.
The total number of words formed using the seven letters from the word CLASSIC is found by using the multiplication principle. There are seven places for each of the seven letters. The first place has 7 choices, the second place has (7-1) =6 choices, the third places has 5 choices and the seventh place has 1 choice. Hence, the total number of words formed is:
= 7*6*5*4 * ... * 1 = 7!
Notice that there are two C's and two S' in the word, which can be treated as repeated elements. To adjust for the repeated elements we will divide 7! by the product of 2!*2!
So, the total number of words formed is:
7!/(2!*2!)
We need to find how many of these words will have the two C's together. To do this, let us treat the two C's as a single entity. So, now we have six spaces to fill. Continuing the same way as in the step above, we can fill the first place in 6 ways, the second place in 5 ways and the sixth place in 1 way. Hence there are 6! ways of forming the words. Once again, we will need to adjust for the two S' which can be done by dividing 6! by 2!.
Total number of 7 lettered words such that the two C's are always together = 6!/2!
The fraction of seven lettered words such that the two C's are always together is:
= (number of words with two C's together/total number of words) = (6!/2!)/(7!/[2!*2!]) = (2!/7) = 2/7
Hence the correct answer choice is A.

Why do we cancel out the two S's and C's? Surely they are separate letters and must be treated as such? I thought the answer was to treat the two C's as one thing and then: 6!/7! = 1/7. I know I am wrong, but don't understand why.
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Re: Why cancel out the 2 S's and C's? (Permutation/Combination) [#permalink] New post 09 Dec 2012, 06:33
jessello wrote:
Here is the question:
What fraction of seven lettered words formed using the letters of the words CLASSIC will have the two C's always together?
A) 2/7
B) 5/7
C) 15/19
D) 4/19
E) 2/8


The answer explanation given is:
This is a typical permutation-probability problem. To make this problem easily understandable we will break into two parts:
i. First, we will find out all the seven lettered words from the letters of word CLASSIC
ii. Next, we will find out how many of these words will have the two C's together.
The total number of words formed using the seven letters from the word CLASSIC is found by using the multiplication principle. There are seven places for each of the seven letters. The first place has 7 choices, the second place has (7-1) =6 choices, the third places has 5 choices and the seventh place has 1 choice. Hence, the total number of words formed is:
= 7*6*5*4 * ... * 1 = 7!
Notice that there are two C's and two S' in the word, which can be treated as repeated elements. To adjust for the repeated elements we will divide 7! by the product of 2!*2!
So, the total number of words formed is:
7!/(2!*2!)
We need to find how many of these words will have the two C's together. To do this, let us treat the two C's as a single entity. So, now we have six spaces to fill. Continuing the same way as in the step above, we can fill the first place in 6 ways, the second place in 5 ways and the sixth place in 1 way. Hence there are 6! ways of forming the words. Once again, we will need to adjust for the two S' which can be done by dividing 6! by 2!.
Total number of 7 lettered words such that the two C's are always together = 6!/2!
The fraction of seven lettered words such that the two C's are always together is:
= (number of words with two C's together/total number of words) = (6!/2!)/(7!/[2!*2!]) = (2!/7) = 2/7
Hence the correct answer choice is A.

Why do we cancel out the two S's and C's? Surely they are separate letters and must be treated as such? I thought the answer was to treat the two C's as one thing and then: 6!/7! = 1/7. I know I am wrong, but don't understand why.


To find all possible words that can be created from the word CLASSIC you must take the number of letters - 7! and then divide it by the repeating letters. you have 2 C's and 2 S's.
The reason they do that is you can have the same word twice, CLASSIC for example, if you switch the 2 C's (or 2 S's) you still get the same word.
so all possibilities are 7!/2!2!.

now, for the second part of the question, try to think about it as if they ask you to seat a few people in a row and 2 people must sit next to each other.
if both C's must stay together, we treat them as if they form 1 letter instead of 2. The 2 C's are attached so lets replace them with * so the new word is -> *LASSI.
now you have 6 letters total -> 6! and 2 repetitions because of the 2 S's so you must divide that by 2!.
finally, you take all possible words where the 2 C's are attached together and divide that by all possible words that form CLASSIC to get the right answer.

hope i was clear enough.
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Re: Why cancel out the 2 S's and C's? (Permutation/Combination) [#permalink] New post 09 Dec 2012, 06:38
I still don't see why we can't have two words that are the same (for example with 2 S's in the above question) - nowhere in the question does it state that we can't. Must I assume that questions mean this?
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Re: Why cancel out the 2 S's and C's? (Permutation/Combination) [#permalink] New post 09 Dec 2012, 06:42
jessello wrote:
I still don't see why we can't have two words that are the same (for example with 2 S's in the above question) - nowhere in the question does it state that we can't. Must I assume that questions mean this?


That's hows Combinatorics work.
I suggest you go through the GMATclub Math Book.
It will make everything clearer.
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Re: Why cancel out the 2 S's and C's? (Permutation/Combination) [#permalink] New post 11 Dec 2012, 05:22
jessello wrote:
Here is the question:
What fraction of seven lettered words formed using the letters of the words CLASSIC will have the two C's always together?
A) 2/7
B) 5/7
C) 15/19
D) 4/19
E) 2/8

Why do we cancel out the two S's and C's? Surely they are separate letters and must be treated as such? I thought the answer was to treat the two C's as one thing and then: 6!/7! = 1/7. I know I am wrong, but don't understand why.


First, if you post in the correct forum you have a chance of better response to your query. In this case: quant P.S.

I'll go ahead and explain:

let us start by an easier example: arrange ILL. so according to the formula you read it should be 3!/2! (because L is repeated twice) and according to you it should be 3! (because the two Ls are separate identities)

cases:
ILL
LIL
LLI

That is all the cases you can make (which is 3!/2! and not 3!)

So why do we divide by 2! when a letter is repeated twice?
basically to negate the double counting of words: if 2 Ls were separate identities L(1)L(2)I and L(2)L(1)I would be two separate words. But they are not. At the end of the day they are just LLI - 1 word. You cannot differentiate between the two Ls when they are together.

Now Why 2! and why not just 2.
Notice that you may arrange LL in two places in 2! ways (and thus by that factor the words will duplicate). 2!=2 so you won't find trouble here. But if the word were MMMI then it would have been 4!/3! and not 4!/3. Because you may arrange MMM in 3! ways.

Makes sense?

You can extend this logic to any number of letters.
Re: Why cancel out the 2 S's and C's? (Permutation/Combination)   [#permalink] 11 Dec 2012, 05:22
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