How many different four letter words can be formed (the : Quant Question Archive [LOCKED]
Check GMAT Club Decision Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 21 Jan 2017, 11:45

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
Your Progress

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 different four letter words can be formed (the

 post reply Question banks Downloads My Bookmarks Reviews Important topics
Author Message
Intern
Joined: 17 Sep 2005
Posts: 7
Location: Delhi
Followers: 0

Kudos [?]: 0 [0], given: 0

How many different four letter words can be formed (the [#permalink]

Show Tags

20 Sep 2005, 22:40
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

How many different four letter words can be formed (the words need not be meaningful) using the letters of the word MEDITERRANEAN such that the first letter is E and the last letter is R?

A. 59
B. 11!/2!*2!*2!
C. 56
D. 23
E. 11!/3!*2!*2!*2!

Please simplify

Thanks
Manager
Joined: 03 Aug 2005
Posts: 134
Followers: 1

Kudos [?]: 2 [0], given: 0

Show Tags

21 Sep 2005, 03:11
I'd go for B

you have 13 letters, two of them are alreadty used, so you have 11! possibilities.

There are two As Ns and Es so you have to divide by 2!*2!*2!
Manager
Joined: 16 Sep 2005
Posts: 62
Followers: 1

Kudos [?]: 2 [0], given: 0

Show Tags

21 Sep 2005, 08:35
I came up with C as an answe. Explaination as follows

1) Ignoring repetitions in all there 8 letters in total.
2) First and last letters are E & R
3) We need to find number of ways middle 2 letter combination can be generated.

As it is not mentioned that E & R can not appear in these two letters I calculated 8P2 (permutation) for an answer.

Please let me know if I am wrong.
_________________

Thanks

VP
Joined: 22 Aug 2005
Posts: 1120
Location: CA
Followers: 1

Kudos [?]: 102 [0], given: 0

Show Tags

21 Sep 2005, 09:05
MBAHopeful wrote:
I came up with C as an answe. Explaination as follows

1) Ignoring repetitions in all there 8 letters in total.
2) First and last letters are E & R
3) We need to find number of ways middle 2 letter combination can be generated.

As it is not mentioned that E & R can not appear in these two letters I calculated 8P2 (permutation) for an answer.

Please let me know if I am wrong.

why are you taking 8 letters only. Isnt 11 letters left to be chosen to fit 2 middle numbers?

the numbers left for middle spots:
E- 2
A - 2
N - 2
R - 1
4 - single occurance letters

we need to find all distinct 2 letter combinations among them. Not sure how to calculated it. Initially i thought it would be:
11p2/2! * 2! * 2!

but calculation results in fraction.
Manager
Joined: 16 Sep 2005
Posts: 62
Followers: 1

Kudos [?]: 2 [0], given: 0

Show Tags

21 Sep 2005, 10:26
duttsit wrote:
MBAHopeful wrote:
I came up with C as an answe. Explaination as follows

1) Ignoring repetitions in all there 8 letters in total.
2) First and last letters are E & R
3) We need to find number of ways middle 2 letter combination can be generated.

As it is not mentioned that E & R can not appear in these two letters I calculated 8P2 (permutation) for an answer.

Please let me know if I am wrong.

why are you taking 8 letters only. Isnt 11 letters left to be chosen to fit 2 middle numbers?

the numbers left for middle spots:
E- 2
A - 2
N - 2
R - 1
4 - single occurance letters

we need to find all distinct 2 letter combinations among them. Not sure how to calculated it. Initially i thought it would be:
11p2/2! * 2! * 2!

but calculation results in fraction.

I thought we need to ignore multiple occurrences of the same letter in the word and hence I came up with number 8. Let me know if my thinking is wrong.
_________________

Thanks

Senior Manager
Joined: 29 Nov 2004
Posts: 484
Location: Chicago
Followers: 1

Kudos [?]: 25 [0], given: 0

Re: P&C [#permalink]

Show Tags

21 Sep 2005, 12:08
I get 59

one E and R are fixed besides that there are 11 letters for two places out of that 8 are distinct so they can be used to produce

8*7 = 56 combinations

plus three letters are repeated twice so they can make 3 more words..

so the answer is 56+3 = 59
_________________

Fear Mediocrity, Respect Ignorance

Manager
Joined: 03 Aug 2005
Posts: 134
Followers: 1

Kudos [?]: 2 [0], given: 0

Show Tags

23 Sep 2005, 00:09
I agree with ranga41, I read the question wrong and thought they were asking for 8 letter words.

Answer is A
Senior Manager
Joined: 08 Sep 2004
Posts: 258
Location: New York City, USA
Followers: 1

Kudos [?]: 17 [0], given: 0

Show Tags

02 Jan 2006, 10:44
I got 59 too. Except I used a different approact.

We have 3 letter with 2 occurence and 5 with single. To pick 2 letters, I can do it in two different ways.

1. pick 3 repeated then any 8.
2. pick 5 single and then any 7.

result 3*8 + 5*7 = 59.

- Vipin
Senior Manager
Joined: 09 Aug 2005
Posts: 285
Followers: 1

Kudos [?]: 3 [0], given: 0

Re: P&C [#permalink]

Show Tags

22 Jan 2006, 09:53
ranga41 wrote:
I get 59

one E and R are fixed besides that there are 11 letters for two places out of that 8 are distinct so they can be used to produce

8*7 = 56 combinations

plus three letters are repeated twice so they can make 3 more words..

so the answer is 56+3 = 59

Can anyone please explain why we are adding 3 here?

The word formed by two 'different E's' will read the same and should be considered as a same word.

Also,

out of 13, 2 are used up E and R

you are left with 11 alphabets with two N's, 2 A's and 2 E's to be filled in 2 places

should it not be

11x10 -3 = 107
words
VP
Joined: 21 Sep 2003
Posts: 1065
Location: USA
Followers: 3

Kudos [?]: 73 [0], given: 0

Show Tags

22 Jan 2006, 12:02
old_dream,

Obviously I too didn't get this right the first time.

But here is what I think:

MEDITERRANEAN has

13 letters with
3 - E's
2 - R's
2 - N's
2 - A's

The words to be formed are of the type E _, _, R

Consider all the single characters withour repititions i.e, we need to choose from : (M, D, I, T, R) + 1 from each of (E, N, A) = 8 characters.

So No of words = 8*7 = 56

Now we had excluded the possibility of (2 Es, 2Ns, 2A's)

So we have only 3 more words to consider:
EAAR
ENNR
EEER

Hence 56+3 = 59
_________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

- Bernard Edmonds

22 Jan 2006, 12:02
Display posts from previous: Sort by

How many different four letter words can be formed (the

 post reply Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.