It is currently 25 Jun 2017, 15:43

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# How many 7 letter words can be constructed using 26 letters

Author Message
Intern
Joined: 26 Mar 2006
Posts: 27
How many 7 letter words can be constructed using 26 letters [#permalink]

### Show Tags

12 Apr 2006, 20:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many 7 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels (repetitions allowed)

a) (21^4)*(5^3)
b) (21^4)*(5^3)*7!
c) {(21^4)*(5^3)*7!}/4!*3!
d) 21C4*5C3*7!
e) None of these
VP
Joined: 29 Dec 2005
Posts: 1341
Re: P&C - 7 letter word [#permalink]

### Show Tags

12 Apr 2006, 21:28
chillpill wrote:
How many 7 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels (repetitions allowed)

a) (21^4)*(5^3)
b) (21^4)*(5^3)*7!
c) {(21^4)*(5^3)*7!}/4!*3!
d) 21C4*5C3*7!
e) None of these

total = 26
vowels (a, e, i, o, and u) = 5
consonants = 21

no of ways 3 vowels can be selected from 5 vowels = 5c3
no of ways 4 (7-3) consonants can be selected from 21 consonants= 21c4
no of ways 7 words can be arranged = 7!
so multiply all then its

(D) 21C4*5C3*7!
Director
Joined: 10 Oct 2005
Posts: 718
Re: P&C - 7 letter word [#permalink]

### Show Tags

12 Apr 2006, 21:39
Professor wrote:
chillpill wrote:
How many 7 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels (repetitions allowed)

a) (21^4)*(5^3)
b) (21^4)*(5^3)*7!
c) {(21^4)*(5^3)*7!}/4!*3!
d) 21C4*5C3*7!
e) None of these

total = 26
vowels (a, e, i, o, and u) = 5
consonants = 21

no of ways 3 vowels can be selected from 5 vowels = 5c3
no of ways 4 (7-3) consonants can be selected from 21 consonants= 21c4
no of ways 7 words can be arranged = 7!
so multiply all then its

(D) 21C4*5C3*7!

10x you are really an El Professore)))
_________________

IE IMBA 2010

Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

### Show Tags

13 Apr 2006, 00:29
Think that since repetitions are allowed it should be B)
Four consonants can be selected each in 21 ways, 3 vowels in 5 ways each and they can have 7! orderings. Now if the words have to be different then we come to A)
Director
Joined: 24 Oct 2005
Posts: 659
Location: London

### Show Tags

13 Apr 2006, 03:17
The 7 letter word can be

V V V C C C C which can be arranged in 7! ways.

Each V can be any of the 5 vowels.
Each C can be any of the 21 Consonants - since repetition is allowed.

(5*5*5 * 21*21*21*21 ) * 7!

Ans = B
Manager
Joined: 09 Feb 2006
Posts: 129
Location: New York, NY

### Show Tags

13 Apr 2006, 07:01
I thought the words do have to be different. This would give us A, correct?
VP
Joined: 28 Mar 2006
Posts: 1369

### Show Tags

13 Apr 2006, 15:25
BG is right...D is correct
VP
Joined: 29 Dec 2005
Posts: 1341

### Show Tags

13 Apr 2006, 17:56
BG wrote:
Think that since repetitions are allowed it should be B)
Four consonants can be selected each in 21 ways, 3 vowels in 5 ways each and they can have 7! orderings. Now if the words have to be different then we come to A)

you are right. i didnot cinsider "repititon."
Senior Manager
Joined: 24 Jan 2006
Posts: 252

### Show Tags

14 Apr 2006, 08:17

(5*5*5 * 21*21*21*21) * 7!
Senior Manager
Joined: 08 Sep 2004
Posts: 257
Location: New York City, USA

### Show Tags

14 Apr 2006, 20:56
Professor wrote:
BG wrote:
Think that since repetitions are allowed it should be B)
Four consonants can be selected each in 21 ways, 3 vowels in 5 ways each and they can have 7! orderings. Now if the words have to be different then we come to A)

you are right. i didnot cinsider "repititon."

But even if you consider repetition, A should not be the answer. I think C is the correct Answer.

- Vipin
14 Apr 2006, 20:56
Display posts from previous: Sort by
The provided text passage is a GMAT Club Daily Prep question post, which includes a problem-solving question related to permutations and combinations. The question asks how many 7-letter words can be constructed using 26 letters of the alphabet, with the condition that each word must contain 3 vowels (repetitions allowed).

### Content Structure:
- The passage starts with a timestamp.
- It then introduces the GMAT Club Daily Prep service.
- It provides a problem-solving question.
- It lists multiple-choice options for the answer.

### Analysis:
single_question_with_answer

the passage contains a single mathematical question and provides multiple-choice answers, including the correct answer. the correct answer is given as (b) (21^4)*(5^3)*7!, which accounts for the selection of 4 consonants (each in 21 ways), 3 vowels (each in 5 ways), and the 7! ways to arrange the 7 letters.
Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
- NO_QUESTION
- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
system
It is currently 25 Jun 2017, 15:43

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# How many 7 letter words can be constructed using 26 letters

Author Message
Intern
Joined: 26 Mar 2006
Posts: 27
How many 7 letter words can be constructed using 26 letters [#permalink]

### Show Tags

12 Apr 2006, 20:09
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many 7 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels (repetitions allowed)

a) (21^4)*(5^3)
b) (21^4)*(5^3)*7!
c) {(21^4)*(5^3)*7!}/4!*3!
d) 21C4*5C3*7!
e) None of these
VP
Joined: 29 Dec 2005
Posts: 1341
Re: P&C - 7 letter word [#permalink]

### Show Tags

12 Apr 2006, 21:28
chillpill wrote:
How many 7 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels (repetitions allowed)

a) (21^4)*(5^3)
b) (21^4)*(5^3)*7!
c) {(21^4)*(5^3)*7!}/4!*3!
d) 21C4*5C3*7!
e) None of these

total = 26
vowels (a, e, i, o, and u) = 5
consonants = 21

no of ways 3 vowels can be selected from 5 vowels = 5c3
no of ways 4 (7-3) consonants can be selected from 21 consonants= 21c4
no of ways 7 words can be arranged = 7!
so multiply all then its

(D) 21C4*5C3*7!
Director
Joined: 10 Oct 2005
Posts: 718
Re: P&C - 7 letter word [#permalink]

### Show Tags

12 Apr 2006, 21:39
Professor wrote:
chillpill wrote:
How many 7 letter words can be constructed using 26 letters of the alphabet if each word contains 3 vowels (repetitions allowed)

a) (21^4)*(5^3)
b) (21^4)*(5^3)*7!
c) {(21^4)*(5^3)*7!}/4!*3!
d) 21C4*5C3*7!
e) None of these

total = 26
vowels (a, e, i, o, and u) = 5
consonants = 21

no of ways 3 vowels can be selected from 5 vowels = 5c3
no of ways 4 (7-3) consonants can be selected from 21 consonants= 21c4
no of ways 7 words can be arranged = 7!
so multiply all then its

(D) 21C4*5C3*7!

10x you are really an El Professore)))
_________________

IE IMBA 2010

Director
Joined: 13 Nov 2003
Posts: 789
Location: BULGARIA

### Show Tags

13 Apr 2006, 00:29
Think that since repetitions are allowed it should be B)
Four consonants can be selected each in 21 ways, 3 vowels in 5 ways each and they can have 7! orderings. Now if the words have to be different then we come to A)
Director
Joined: 24 Oct 2005
Posts: 659
Location: London

### Show Tags

13 Apr 2006, 03:17
The 7 letter word can be

V V V C C C C which can be arranged in 7! ways.

Each V can be any of the 5 vowels.
Each C can be any of the 21 Consonants - since repetition is allowed.

(5*5*5 * 21*21*21*21 ) * 7!

Ans = B
Manager
Joined: 09 Feb 2006
Posts: 129
Location: New York, NY

### Show Tags

13 Apr 2006, 07:01
I thought the words do have to be different. This would give us A, correct?
VP
Joined: 28 Mar 2006
Posts: 1369

### Show Tags

13 Apr 2006, 15:25
BG is right...D is correct
VP
Joined: 29 Dec 2005
Posts: 1341

### Show Tags

13 Apr 2006, 17:56
BG wrote:
Think that since repetitions are allowed it should be B)
Four consonants can be selected each in 21 ways, 3 vowels in 5 ways each and they can have 7! orderings. Now if the words have to be different then we come to A)

you are right. i didnot cinsider "repititon."
Senior Manager
Joined: 24 Jan 2006
Posts: 252

### Show Tags

14 Apr 2006, 08:17

(5*5*5 * 21*21*21*21) * 7!
Senior Manager
Joined: 08 Sep 2004
Posts: 257
Location: New York City, USA

### Show Tags

14 Apr 2006, 20:56
Professor wrote:
BG wrote:
Think that since repetitions are allowed it should be B)
Four consonants can be selected each in 21 ways, 3 vowels in 5 ways each and they can have 7! orderings. Now if the words have to be different then we come to A)

you are right. i didnot cinsider "repititon."

But even if you consider repetition, A should not be the answer. I think C is the correct Answer.

- Vipin
14 Apr 2006, 20:56
Display posts from previous: Sort by
user
The provided text passage is a GMAT Club Daily Prep question post, which includes a problem-solving question related to permutations and combinations. The question asks how many 7-letter words can be constructed using 26 letters of the alphabet, with the condition that each word must contain 3 vowels (repetitions allowed).

### Content Structure:
- The passage starts with a timestamp.
- It then introduces the GMAT Club Daily Prep service.
- It provides a problem-solving question.
- It lists multiple-choice options for the answer.

### Analysis:
TYPE: SINGLE_QUESTION_WITH_ANSWER

The passage contains a single mathematical question and provides multiple-choice answers, including the correct answer. The correct answer is given as (B) (21^4)*(5^3)*7!, which accounts for the selection of 4 consonants (each in 21 ways), 3 vowels (each in 5 ways), and the 7! ways to arrange the 7 letters.