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How many words can be formed by taking 4 letters at a time

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How many words can be formed by taking 4 letters at a time [#permalink] New post 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] New post 14 Apr 2010, 06:45
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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: M-A-T-H-E-I-C-S. 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] New post 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] New post 14 Apr 2010, 07:48
Bunuel wrote:
3. aabc - from 4 letters 2 are the same and other 2 are different:
3C1*7C1*6C1*\frac{4!}{2!}=756 - 3C1 choosing which letter will proved with 2 letters (out of 3 double letter - MAT), 7C1 choosing third letter out of 7 distinct letters left, 6C1 choosing fourth letter out of 6 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).




M-A-T-H-E-I-C-S
M-A-T

Y is it 7C1*6C1? selecting 2 from 7 is 7C2?....
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Re: tough p n c [#permalink] New post 14 Apr 2010, 08:01
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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
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Re: tough p n c [#permalink] New post 14 Apr 2010, 08:03
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RaviChandra wrote:
Bunuel wrote:
3. aabc - from 4 letters 2 are the same and other 2 are different:
3C1*7C1*6C1*\frac{4!}{2!}=756 - 3C1 choosing which letter will proved with 2 letters (out of 3 double letter - MAT), 7C1 choosing third letter out of 7 distinct letters left, 6C1 choosing fourth letter out of 6 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).




M-A-T-H-E-I-C-S
M-A-T

Y is it 7C1*6C1? selecting 2 from 7 is 7C2?....


It's a typo. There should be 7C1*6C1/2, which is in fact 7C2. Edited.
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PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!! ,11 Mixed Questions NEW!!!, 12 Fresh Meat NEW!!!

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Re: tough p n c [#permalink] New post 14 Apr 2010, 11:41
thanks a ton, bunuel :)
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Re: tough p n c [#permalink] New post 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] New post 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: M-A-T-H-E-I-C-S. 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] New post 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: M-A-T-H-E-I-C-S. 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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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. NEW!!! ,11 Mixed Questions NEW!!!, 12 Fresh Meat NEW!!!

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set. NEW!!!, 11 New DS set. NEW!!!


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Re: tough p n c   [#permalink] 26 May 2013, 04:13
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