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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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thanks a ton, bunuel :)
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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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Bunuel
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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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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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Bunuel
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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.

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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Bunuel
jatt86
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.

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: best-gmat-math-prep-books-reviews-recommendations-77291.html

Theory on Combinations: math-combinatorics-87345.html

DS questions on Combinations: search.php?search_id=tag&tag_id=31
PS questions on Combinations: search.php?search_id=tag&tag_id=52

Tough and tricky questions on Combinations: hardest-area-questions-probability-and-combinations-101361.html

Hope it helps.
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If this would have been a word with three of the same letter I'm assuming you would have more than 3 combinations?

Thanks!
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If this would have been a word with three of the same letter I'm assuming you would have more than 3 combinations?

Thanks!

Yes, we would have one more combination {a, a, a, b}.
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I can't understand this question. Why is this a combination question and not permutation? isnt it asking for arrangements?
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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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Thanks! :) this helped :)
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jatt86
How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS.

A. 756
B. 1680
C. 1698
D. 2436
E. 2454

Asked: How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS.

M-2
A-2
T-2
H-1
E-1
I-1
C-1
S-1

Words of the form abcd = \(^8C_4 * 4! = 1680\)

Words of the form aabc = \(^3C_1*^7C_2* 4!/2! = 3*21*12 = 756\)

Words of the form aabb = \(^3C_2 * 4!/2!/2! = 3* 24/4 = 18\)

Number of words that can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS = 1680+756+18=2454

IMO E
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I understand the solution, but where does this logic break down?
MATHEMATICS = 11 letters
MM AA TT = 3 groups of 2 repeats

(11*10*9*8)/(2!2!2!) = 990
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Bunuel
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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.


hey! Why are we taking 4C2 in the aabb combination? If we are looking to calculate how the letters are arranged, shouldn't we be using 4P2 instead?

Thanks :)
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Bunuel
jatt86
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.


Bunuel Could you please help me understand why did we take 3C2 and 3C1 and 7C2?
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Bunuel
jatt86
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.


Bunuel Could you please help me understand why did we take 3C2 and 3C1 and 7C2?

"aabb" case means that there are two pairs of identical letters, with one pair being "a" and the other pair being "b". Now, we have three pairs of identical letters in MATHEMATICS: {MM}, {AA} and {TT}. There are 3C2 =3 ways to choose two pairs out of three:
{MM}, {AA};
{MM}, {TT};
{AA}, {TT}.
In the next step we are arranging four letters, where two pairs are identical with 4!/(2!2!).

"aabc" case means that there are a pair of identical letters, and two distinct letters. There are 3C1 = 3 ways to choose a pair of double letters out of three:
{MM},
{AA};
{TT}.
Then we need two distinct letters. Say we chose {MM} for a pair of letter. We'd be left with the following 7 letters: A-T-H-E-I-C-S. The number of ways to choose 2 letters out of 7 is 7C2. Finally, we need to arrange this case with 4!/2!.

Hope it's clear.
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