Oct 16 08:00 PM PDT  09:00 PM PDT EMPOWERgmat is giving away the complete Official GMAT Exam Pack collection worth $100 with the 3 Month Pack ($299) Oct 18 08:00 AM PDT  09:00 AM PDT Learn an intuitive, systematic approach that will maximize your success on Fillintheblank GMAT CR Questions. Oct 19 07:00 AM PDT  09:00 AM PDT Does GMAT RC seem like an uphill battle? eGMAT is conducting a free webinar to help you learn reading strategies that can enable you to solve 700+ level RC questions with at least 90% accuracy in less than 10 days. Sat., Oct 19th at 7 am PDT Oct 20 07:00 AM PDT  09:00 AM PDT Get personalized insights on how to achieve your Target Quant Score. Oct 22 08:00 PM PDT  09:00 PM PDT On Demand for $79. For a score of 4951 (from current actual score of 40+) AllInOne Standard & 700+ Level Questions (150 questions)
Author 
Message 
TAGS:

Hide Tags

Intern
Joined: 11 Sep 2011
Posts: 9

How many words of 4 letters can be formed from the word "MED
[#permalink]
Show Tags
13 Sep 2011, 10:08
Question Stats:
50% (00:47) correct 50% (04:31) wrong based on 6 sessions
HideShow timer Statistics
How many words of 4 letters can be formed from the word "MEDITERRANEAN"?



Senior Manager
Joined: 11 May 2011
Posts: 295

Re: Permutation...
[#permalink]
Show Tags
13 Sep 2011, 10:17
samcot wrote: Please guide me to how to get the answer for the below question..
How many words of 4 letters can be formed from the word "MEDITERRANEAN"?
thanks in advance!!! Total number of letters = 13, Repetations  3E, 2R, 2N, 2A Answer  13C4/(3!*2!*2!*2!) Cheers!
_________________
 What you do TODAY is important because you're exchanging a day of your life for it! 



Intern
Joined: 11 Sep 2011
Posts: 9

Re: Permutation...
[#permalink]
Show Tags
13 Sep 2011, 10:47
Ajay369 wrote: samcot wrote: Please guide me to how to get the answer for the below question..
How many words of 4 letters can be formed from the word "MEDITERRANEAN"?
thanks in advance!!! Total number of letters = 13, Repetations  3E, 2R, 2N, 2A Answer  13C4/(3!*2!*2!*2!) Cheers! 13C4/(3!*2!*2!*2!)= 14.895... can the no of words be negative?? This approach is wrong i think since it is not a combination problem.. the approach to the solution should be sth like this : no of words without repetition = 8*7*6*5 =1680 + no of words with repetition=???? or no of words without considering the repetition= 13P4  no of words that are repeated thanks in advance!!!



Senior Manager
Joined: 11 May 2011
Posts: 295

Re: Permutation...
[#permalink]
Show Tags
13 Sep 2011, 11:06
@samcot  Yes...I know its incorrect. Typed during my meeting without thinking much Will respond again Sorry for inconvenience. Cheers!
_________________
 What you do TODAY is important because you're exchanging a day of your life for it! 



Intern
Joined: 09 Jun 2011
Posts: 18

Re: Permutation...
[#permalink]
Show Tags
13 Sep 2011, 11:49
samcot wrote: Please guide me to how to get the answer for the below question..
How many words of 4 letters can be formed from the word "MEDITERRANEAN"?
thanks in advance!!! 13 letters, 3e,2r,2a,2n,m,d,i,t.. if all 4 different: 8C4*4! if 2 different and one repeat (abca) > 4C1*7C2* (4!/2!) if 2 same and other 2 same (aabb) > 4C2* (4!/2!) if 3 same, 1 different:(aaab) 1C1*7C1*(4!/3!) if all four same: 0 add all....



Senior Manager
Joined: 11 May 2011
Posts: 295

Re: Permutation...
[#permalink]
Show Tags
13 Sep 2011, 12:14
naveen1003 wrote: samcot wrote: Please guide me to how to get the answer for the below question..
How many words of 4 letters can be formed from the word "MEDITERRANEAN"?
thanks in advance!!! 13 letters, 3e,2r,2a,2n,m,d,i,t.. if all 4 different: 8C4*4! if 2 different and one repeat (abca) > 4C1*7C2* (4!/2!) if 2 same and other 2 same (aabb) > 4C2* (4!/2![highlight]*2![/highlight]) if 3 same, 1 different:(aaab) 1C1*7C1*(4!/3!) if all four same: 0 add all.... Looks correct! only one above correction.
_________________
 What you do TODAY is important because you're exchanging a day of your life for it! 



Intern
Joined: 11 Sep 2011
Posts: 9

Re: Permutation...
[#permalink]
Show Tags
13 Sep 2011, 20:52
naveen1003 wrote: samcot wrote: Please guide me to how to get the answer for the below question..
How many words of 4 letters can be formed from the word "MEDITERRANEAN"?
thanks in advance!!! 13 letters, 3e,2r,2a,2n,m,d,i,t.. if all 4 different: 8C4*4! if 2 different and one repeat (abca) > 4C1*7C2* (4!/2!) if 2 same and other 2 same (aabb) > 4C2* (4!/2!) if 3 same, 1 different:(aaab) 1C1*7C1*(4!/3!) if all four same: 0 add all.... I always try to solve this kind of problem with permutation... if all 4 different: 8C4*4! this I understood ,as 8C4*4!= 8P4 where as i am not being able to interpret the followings: when repetition is there if 2 different and one repeat (abca) > 4C1*7C2* (4!/2!) if 2 same and other 2 same (aabb) > 4C2* (4!/2!) It would be of great help if you can elucidate on the cases of repetition.Thanks in advance!!!



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9705
Location: Pune, India

Re: Permutation...
[#permalink]
Show Tags
13 Sep 2011, 21:41
samcot wrote: Please guide me to how to get the answer for the below question..
How many words of 4 letters can be formed from the word "MEDITERRANEAN"?
thanks in advance!!! You cannot use a single step permutation here. You will need to first select and then arrange since selection will vary in each case. We have 8 distinct letters: M, E, D, I, T, R, A, N Then there are some repetitions: 3E, 2R, 2A, 2N In how many ways can you make a 4 letter word? Case 1: All different letters From the 8 distinct letters, you choose 4 and arrange them. = 8C4 * 4! Case 2: 2 letters same, others different For the 2 same letters, choose one from the 4 which are repeated in 4C1 ways. Then to choose 2 other letters, pick two from the rest 7 distinct letters in 7C2 ways. Then arrange them in 4!/2! ways (you divide by 2! because one letter is repeated) = 4C1 * 7C2 * 4!/2! Case 3: 2 letters same, 2 letters same Choose 2 letters from the 4 which are repeated in 4C2 ways. Then arrange them in 4!/(2!*2!) ways (2 letters are repeated so you divide by 2! twice) = 4C2 * 4!/(2!*2!) Case 4: 3 letters same, fourth different Only 'E' appears 3 times so E must be chosen. You can choose the fourth letter from the other 7 letters in 7C1 ways. Arrange them in 4!/3! ways = 7C1 * 4!/3! All four letters cannot be the same since no letter appears four times. To get the final answer, we will need to add the result from all the cases. I wouldn't worry about doing it. This isn't a GMAT type question. Needs a long monotonous approach. GMAT questions can be solved quickly and usually have a trick. This question is useful only to help you understand the basics of permutation and combination.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 11 Sep 2011
Posts: 9

Re: Permutation...
[#permalink]
Show Tags
13 Sep 2011, 22:53
VeritasPrepKarishma wrote: samcot wrote: Please guide me to how to get the answer for the below question..
How many words of 4 letters can be formed from the word "MEDITERRANEAN"?
thanks in advance!!! You cannot use a single step permutation here. You will need to first select and then arrange since selection will vary in each case. We have 8 distinct letters: M, E, D, I, T, R, A, N Then there are some repetitions: 3E, 2R, 2A, 2N In how many ways can you make a 4 letter word? Case 1: All different letters From the 8 distinct letters, you choose 4 and arrange them. = 8C4 * 4! Case 2: 2 letters same, others different For the 2 same letters, choose one from the 4 which are repeated in 4C1 ways. Then to choose 2 other letters, pick two from the rest 7 distinct letters in 7C2 ways. Then arrange them in 4!/2! ways (you divide by 2! because one letter is repeated) = 4C1 * 7C2 * 4!/2! Case 3: 2 letters same, 2 letters same Choose 2 letters from the 4 which are repeated in 4C2 ways. Then arrange them in 4!/(2!*2!) ways (2 letters are repeated so you divide by 2! twice) = 4C2 * 4!/(2!*2!) Case 4: 3 letters same, fourth different Only 'E' appears 3 times so E must be chosen. You can choose the fourth letter from the other 7 letters in 7C1 ways. Arrange them in 4!/3! ways = 7C1 * 4!/3! All four letters cannot be the same since no letter appears four times. To get the final answer, we will need to add the result from all the cases. I wouldn't worry about doing it. This isn't a GMAT type question. Needs a long monotonous approach. GMAT questions can be solved quickly and usually have a trick. This question is useful only to help you understand the basics of permutation and combination. Case 2: 2 letters same, others different For the 2 same letters, choose one from the 4 which are repeated in 4C1 ways. Then to choose 2 other letters, pick two from the rest 7 distinct letters in 7C2 ways. Then arrange them in 4!/2! ways (you divide by 2! because one letter is repeated) = 4C1 * 7C2 * 4!/2! Why it is not 4C1*1! * 7C2*2! *4!/2! Since 7C2 can be arranged themselves in 2! ways ..rt? Let me know what i am missing Thanks!!!



Manager
Status: Bell the GMAT!!!
Affiliations: Aidha
Joined: 16 Aug 2011
Posts: 127
Location: Singapore
Concentration: Finance, General Management
GMAT 1: 680 Q46 V37 GMAT 2: 620 Q49 V27 GMAT 3: 700 Q49 V36
WE: Other (Other)

Re: Permutation...
[#permalink]
Show Tags
14 Sep 2011, 00:17
Quote: Case 1: All different letters From the 8 distinct letters, you choose 4 and arrange them. = 8C4 * 4! Hi Karishma, Can you help me understanding why are we multiplying 8C4 by 4!. Thanks.
_________________
If my post did a dance in your mind, send me the steps through kudos :)
My MBA journey at http://mbadilemma.wordpress.com/



Manager
Status: Bell the GMAT!!!
Affiliations: Aidha
Joined: 16 Aug 2011
Posts: 127
Location: Singapore
Concentration: Finance, General Management
GMAT 1: 680 Q46 V37 GMAT 2: 620 Q49 V27 GMAT 3: 700 Q49 V36
WE: Other (Other)

Re: Permutation...
[#permalink]
Show Tags
14 Sep 2011, 00:21
GMATmission wrote: Quote: Case 1: All different letters From the 8 distinct letters, you choose 4 and arrange them. = 8C4 * 4! Hi Karishma, Can you help me understanding why are we multiplying 8C4 by 4!. Thanks. Got it. Silly question!
_________________
If my post did a dance in your mind, send me the steps through kudos :)
My MBA journey at http://mbadilemma.wordpress.com/



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 9705
Location: Pune, India

Re: Permutation...
[#permalink]
Show Tags
14 Sep 2011, 22:13
samcot wrote: Case 2: 2 letters same, others different For the 2 same letters, choose one from the 4 which are repeated in 4C1 ways. Then to choose 2 other letters, pick two from the rest 7 distinct letters in 7C2 ways. Then arrange them in 4!/2! ways (you divide by 2! because one letter is repeated) = 4C1 * 7C2 * 4!/2!
Why it is not 4C1*1! * 7C2*2! *4!/2! Since 7C2 can be arranged themselves in 2! ways ..rt?
Let me know what i am missing
Thanks!!! Think of it this way: You have lots of letters. You need to make a four letter word. How will you do it? You will select 4 letters and then arrange the 4 of them in different ways to get different words. How do you select the four letters? You say 4C1 (to get the one which is repeated) * 7C2 (any two of the remaining 7) Your selection is done. You have 4 letters. Now you want to arrange them. That is done in 4!/2! ways. This includes arranging the 2 distinct letters and the two same ones together. You don't need to arrange the distinct letters separately. This number includes combinations such as EEMD and EEDM.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >



Intern
Joined: 11 Sep 2011
Posts: 9

Re: Permutation...
[#permalink]
Show Tags
17 Sep 2011, 06:39
VeritasPrepKarishma wrote: samcot wrote: Case 2: 2 letters same, others different For the 2 same letters, choose one from the 4 which are repeated in 4C1 ways. Then to choose 2 other letters, pick two from the rest 7 distinct letters in 7C2 ways. Then arrange them in 4!/2! ways (you divide by 2! because one letter is repeated) = 4C1 * 7C2 * 4!/2!
Why it is not 4C1*1! * 7C2*2! *4!/2! Since 7C2 can be arranged themselves in 2! ways ..rt?
Let me know what i am missing
Thanks!!! Think of it this way: You have lots of letters. You need to make a four letter word. How will you do it? You will select 4 letters and then arrange the 4 of them in different ways to get different words. How do you select the four letters? You say 4C1 (to get the one which is repeated) * 7C2 (any two of the remaining 7) Your selection is done. You have 4 letters. Now you want to arrange them. That is done in 4!/2! ways. This includes arranging the 2 distinct letters and the two same ones together. You don't need to arrange the distinct letters separately. This number includes combinations such as EEMD and EEDM. Loads of Thanks to u,Karishma..



Manager
Joined: 10 Jan 2011
Posts: 114
Location: India
GMAT Date: 07162012
GPA: 3.4
WE: Consulting (Consulting)

Re: Permutation...
[#permalink]
Show Tags
30 Sep 2011, 03:55
I don't think this type of question will appear on GMAT.... any very good qestion to understand the difference between permutation and combination
_________________
Analyze why option A in SC wrong



NonHuman User
Joined: 09 Sep 2013
Posts: 13203

Re: How many words of 4 letters can be formed from the word "MED
[#permalink]
Show Tags
29 Jan 2019, 19:36
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________




Re: How many words of 4 letters can be formed from the word "MED
[#permalink]
29 Jan 2019, 19:36






