How many words can be formed by taking 4 letters at a time of the word ineffective?

Sum

How many 4 letter words can be formed using letters in the word MADHURI if letters cannot be repeated?

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Solution

When repetition of letters is not allowed, the number of 4-letter words formed from the letters of the word MADHURI is

∴ 7P4 = `(7!)/((7-4)!)=(7xx6xx5xx4xx3!)/(3!)` = 840
∴ 840 four-letter words can be formed when the repetition of letters is not allowed.

Concept: Permutations - Permutations When All Objects Are Distinct

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Chapter 6: Permutations and Combinations - Exercise 6.3 [Page 81]

Q 5. (ii)Q 5. (i)Q 6. (i)

APPEARS IN

Balbharati Mathematics and Statistics 2 (Commerce) 11th Standard HSC Maharashtra State Board

Chapter 6 Permutations and Combinations
Exercise 6.3 | Q 5. (ii) | Page 81

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How many words can be formed by taking 4 letters at a time out of the [#permalink]

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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.

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

Originally posted by jatt86 on 14 Apr 2010, 04:33.
Last edited by Bunuel on 26 Feb 2019, 03:23, edited 1 time in total.

Added options.

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Joined: 02 Sep 2009

Posts: 86665

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

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

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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Joined: 02 Sep 2009

Posts: 86665

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

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

thanks a ton, bunuel

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Joined: 21 Mar 2010

Posts: 94

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

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

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?

Math Expert

Joined: 02 Sep 2009

Posts: 86665

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

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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Posts: 8

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

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.

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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Joined: 02 Sep 2009

Posts: 86665

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

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

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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Posts: 86665

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

Mbearmann wrote:

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

I can't understand this question. Why is this a combination question and not permutation? isnt it asking for arrangements?

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

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

Thanks!

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

jatt86 wrote:

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

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

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.

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

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

27 Feb 2020, 17:50

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