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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).

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).

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

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.
_________________

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!

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).

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.
_________________

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?

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.

Re: How many words can be formed by taking 4 letters at a time [#permalink]

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20 Apr 2015, 15:53

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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.
_________________

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?

gmatclubot

Re: How many words can be formed by taking 4 letters at a time
[#permalink]
28 Nov 2015, 12:14

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