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How many words can be formed by taking 4 letters at a time [#permalink]
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14 Apr 2010, 05:33
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How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS.



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Re: tough p n c [#permalink]
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jatt86 wrote: 1) how many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. There are 8 distinct letters: MATHEICS. 3 letters M, A, and T are represented twice (double letter). Selected 4 letters can have following 3 patterns: 1. abcd  all 4 letters are different: \(8P4=1680\) (choosing 4 distinct letters out of 8, when order matters) or \(8C4*4!=1680\) (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters); 2. aabb  from 4 letters 2 are the same and other 2 are also the same: \(3C2*\frac{4!}{2!2!}=18\)  3C2 choosing which two double letter will provide two letters (out of 3 double letter  MAT), multiplying by \(\frac{4!}{2!2!}\) to get different arrangements (for example MMAA can be arranged in \(\frac{4!}{2!2!}\) # of ways); 3. aabc  from 4 letters 2 are the same and other 2 are different: \(3C1*7C2*\frac{4!}{2!}=756\)  3C1 choosing which letter will proved with 2 letters (out of 3 double letter  MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by \(\frac{4!}{2!}\) to get different arrangements (for example MMIC can be arranged in \(\frac{4!}{2!}\) # of ways). 1680+18+756=2454 Answer: 2454.
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Re: tough p n c [#permalink]
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14 Apr 2010, 07:40
a very similar question: Find the no: of 4 letter words that can be formed from the string "AABBBBCC" ? Here we have 3 distinct letters(A,B,C) & 4 slots to fill. What logic do you use to solve this problem?
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Re: tough p n c [#permalink]
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14 Apr 2010, 11:41
thanks a ton, bunuel
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Re: tough p n c [#permalink]
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15 Apr 2010, 07:20
I'm usually not bad with anagram problems like this but the term "words" threw me off completely. For some reason I assumed the combination of letters had to combine to make sense, i.e. a "word".
MTHE  is hardly a word, so i started counting actual "words"... so obviously completely bombed the question!



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Re: tough p n c [#permalink]
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25 May 2013, 18:50
Bunuel wrote: jatt86 wrote: 1) how many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. There are 8 distinct letters: MATHEICS. 3 letters M, A, and T are represented twice (double letter). Selected 4 letters can have following 3 patterns: 1. abcd  all 4 letters are different: \(8P4=1680\) (choosing 4 distinct letters out of 8, when order matters) or \(8C4*4!=1680\) (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters); 2. aabb  from 4 letters 2 are the same and other 2 are also the same: \(3C2*\frac{4!}{2!2!}=18\)  3C2 choosing which two double letter will provide two letters (out of 3 double letter  MAT), multiplying by \(\frac{4!}{2!2!}\) to get different arrangements (for example MMAA can be arranged in \(\frac{4!}{2!2!}\) # of ways); 3. aabc  from 4 letters 2 are the same and other 2 are different: \(3C1*7C2*\frac{4!}{2!}=756\)  3C1 choosing which letter will proved with 2 letters (out of 3 double letter  MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by \(\frac{4!}{2!}\) to get different arrangements (for example MMIC can be arranged in \(\frac{4!}{2!}\) # of ways). 1680+18+756=2454 Answer: 2454. Hi Bunnel, Is this a GMAT worthy question?



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Re: tough p n c [#permalink]
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26 May 2013, 04:13
cumulonimbus wrote: Bunuel wrote: jatt86 wrote: 1) how many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. There are 8 distinct letters: MATHEICS. 3 letters M, A, and T are represented twice (double letter). Selected 4 letters can have following 3 patterns: 1. abcd  all 4 letters are different: \(8P4=1680\) (choosing 4 distinct letters out of 8, when order matters) or \(8C4*4!=1680\) (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters); 2. aabb  from 4 letters 2 are the same and other 2 are also the same: \(3C2*\frac{4!}{2!2!}=18\)  3C2 choosing which two double letter will provide two letters (out of 3 double letter  MAT), multiplying by \(\frac{4!}{2!2!}\) to get different arrangements (for example MMAA can be arranged in \(\frac{4!}{2!2!}\) # of ways); 3. aabc  from 4 letters 2 are the same and other 2 are different: \(3C1*7C2*\frac{4!}{2!}=756\)  3C1 choosing which letter will proved with 2 letters (out of 3 double letter  MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by \(\frac{4!}{2!}\) to get different arrangements (for example MMIC can be arranged in \(\frac{4!}{2!}\) # of ways). 1680+18+756=2454 Answer: 2454. Hi Bunnel, Is this a GMAT worthy question? No, but this question is good to practice.
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Re: tough p n c [#permalink]
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02 Jul 2013, 11:45
Bunuel wrote: jatt86 wrote: 1) how many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. There are 8 distinct letters: MATHEICS. 3 letters M, A, and T are represented twice (double letter). Selected 4 letters can have following 3 patterns: 1. abcd  all 4 letters are different: \(8P4=1680\) (choosing 4 distinct letters out of 8, when order matters) or \(8C4*4!=1680\) (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters); 2. aabb  from 4 letters 2 are the same and other 2 are also the same: \(3C2*\frac{4!}{2!2!}=18\)  3C2 choosing which two double letter will provide two letters (out of 3 double letter  MAT), multiplying by \(\frac{4!}{2!2!}\) to get different arrangements (for example MMAA can be arranged in \(\frac{4!}{2!2!}\) # of ways); 3. aabc  from 4 letters 2 are the same and other 2 are different: \(3C1*7C2*\frac{4!}{2!}=756\)  3C1 choosing which letter will proved with 2 letters (out of 3 double letter  MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by \(\frac{4!}{2!}\) to get different arrangements (for example MMIC can be arranged in \(\frac{4!}{2!}\) # of ways). 1680+18+756=2454 Answer: 2454. Bunuel, this is a damn hard question and I find myself not fully able to understand your logic. I am from a very weak background but I have poured through all of the MGMAT math books (excluding the advanced one) several times and still find myself unable to intutively figure out the steps to this problem. What extra review would you suggest so I can be able to at least follow your solutions to these answers?



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Re: tough p n c [#permalink]
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02 Jul 2013, 12:21
tmipanthers wrote: Bunuel wrote: jatt86 wrote: 1) how many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. There are 8 distinct letters: MATHEICS. 3 letters M, A, and T are represented twice (double letter). Selected 4 letters can have following 3 patterns: 1. abcd  all 4 letters are different: \(8P4=1680\) (choosing 4 distinct letters out of 8, when order matters) or \(8C4*4!=1680\) (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters); 2. aabb  from 4 letters 2 are the same and other 2 are also the same: \(3C2*\frac{4!}{2!2!}=18\)  3C2 choosing which two double letter will provide two letters (out of 3 double letter  MAT), multiplying by \(\frac{4!}{2!2!}\) to get different arrangements (for example MMAA can be arranged in \(\frac{4!}{2!2!}\) # of ways); 3. aabc  from 4 letters 2 are the same and other 2 are different: \(3C1*7C2*\frac{4!}{2!}=756\)  3C1 choosing which letter will proved with 2 letters (out of 3 double letter  MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by \(\frac{4!}{2!}\) to get different arrangements (for example MMIC can be arranged in \(\frac{4!}{2!}\) # of ways). 1680+18+756=2454 Answer: 2454. Bunuel, this is a damn hard question and I find myself not fully able to understand your logic. I am from a very weak background but I have poured through all of the MGMAT math books (excluding the advanced one) several times and still find myself unable to intutively figure out the steps to this problem. What extra review would you suggest so I can be able to at least follow your solutions to these answers? This question is out of the scope of the GMAT, so I wouldn't worry about it too much. As for the recommendations. Best GMAT Books: bestgmatmathprepbooksreviewsrecommendations77291.htmlTheory on Combinations: mathcombinatorics87345.htmlDS questions on Combinations: search.php?search_id=tag&tag_id=31PS questions on Combinations: search.php?search_id=tag&tag_id=52Tough and tricky questions on Combinations: hardestareaquestionsprobabilityandcombinations101361.htmlHope it helps.
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Re: How many words can be formed by taking 4 letters at a time [#permalink]
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24 Mar 2014, 13:36
If this would have been a word with three of the same letter I'm assuming you would have more than 3 combinations?
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10 Jul 2015, 11:44
I can't understand this question. Why is this a combination question and not permutation? isnt it asking for arrangements?



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10 Jul 2015, 12:03
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ddg wrote: I can't understand this question. Why is this a combination question and not permutation? isnt it asking for arrangements? You are half right. Permutations = Combinations * n! (where n is the number of 'elements'). In this question, you first need to select the letters out of the given one (combination implied as selection = combination!!) and only after you have selected the letters , you can look at the arrangements. You can not directly go to arrangements as you need to follow the 2 step process: 1. Choose 4 out of 11 letters 2. Arrangement of those selections of 4 letters to get all the possible arrangements. Your approach would have been correct, had the question ask us to arrange all of these 11 letters into words of 11 letters or if all the letters were different.



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10 Jul 2015, 13:24
Thanks! this helped



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Re: How many words can be formed by taking 4 letters at a time [#permalink]
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28 Nov 2015, 12:14
Bunuel wrote: idiot wrote: a very similar question: Find the no: of 4 letter words that can be formed from the string "AABBBBCC" ?
Here we have 3 distinct letters(A,B,C) & 4 slots to fill. What logic do you use to solve this problem? Three patterns: 1. XXXX  only BBBB, so 1 2. XXYY  3C2(choosing which will take the places of X and Y from A, B and C)*4!/2!2!(arranging)=18 3. XXYZ  3C1(choosing which will take the place of X from A, B and C)*4!/2!(arranging)=36 4. XXXY  2C1(choosing which will take the place of Y from A and C, as X can be only B)*4!/3!(arranging)=8 1+18+36+8=63 In 3, through 3C1 we choose 1 group of letters from A, B, and C, which will take the place of XX, but what about the remaining 2 letters YZ? One more query, if through 3C1 we choose group B, then we have 4 letters in hand unlike A and C. Doesn't it require to consider?
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Re: How many words can be formed by taking 4 letters at a time [#permalink]
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06 Dec 2017, 12:36
Bunuel wrote: jatt86 wrote: 1) how many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS. There are 8 distinct letters: MATHEICS. 3 letters M, A, and T are represented twice (double letter). Selected 4 letters can have following 3 patterns: 1. abcd  all 4 letters are different: \(8P4=1680\) (choosing 4 distinct letters out of 8, when order matters) or \(8C4*4!=1680\) (choosing 4 distinct letters out of 8 when order does not matter and multiplying by 4! to get different arrangement of these 4 distinct letters); 2. aabb  from 4 letters 2 are the same and other 2 are also the same: \(3C2*\frac{4!}{2!2!}=18\)  3C2 choosing which two double letter will provide two letters (out of 3 double letter  MAT), multiplying by \(\frac{4!}{2!2!}\) to get different arrangements (for example MMAA can be arranged in \(\frac{4!}{2!2!}\) # of ways); 3. aabc  from 4 letters 2 are the same and other 2 are different: \(3C1*7C2*\frac{4!}{2!}=756\)  3C1 choosing which letter will proved with 2 letters (out of 3 double letter  MAT), 7C2 choosing third and fourth letters out of 7 distinct letters left and multiplying by \(\frac{4!}{2!}\) to get different arrangements (for example MMIC can be arranged in \(\frac{4!}{2!}\) # of ways). 1680+18+756=2454 Answer: 2454. In (1) why you use order matters or doesn't matter to solve it? To saw two different ways to find the solution?




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