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# How many words of 4 letters can be formed from the word "MED

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How many words of 4 letters can be formed from the word "MED [#permalink]  13 Sep 2011, 09:08
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100% (01:47) correct 0% (00:00) wrong based on 2 sessions
How many words of 4 letters can be formed from the word "MEDITERRANEAN"?
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Re: Permutation... [#permalink]  13 Sep 2011, 09:17
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"?

Total number of letters = 13, Repetations - 3E, 2R, 2N, 2A

Cheers!
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Re: Permutation... [#permalink]  13 Sep 2011, 09:47
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"?

Total number of letters = 13, Repetations - 3E, 2R, 2N, 2A

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

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Re: Permutation... [#permalink]  13 Sep 2011, 10:06
@samcot - Yes...I know its incorrect. Typed during my meeting without thinking much
Will respond again

Sorry for inconvenience.
Cheers!
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Re: Permutation... [#permalink]  13 Sep 2011, 10:49
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"?

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
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Re: Permutation... [#permalink]  13 Sep 2011, 11:14
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"?

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

Looks correct! only one above correction.
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Re: Permutation... [#permalink]  13 Sep 2011, 19:52
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"?

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

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.

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Re: Permutation... [#permalink]  13 Sep 2011, 20:41
Expert's post
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"?

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.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 11 Sep 2011 Posts: 9 Followers: 0 Kudos [?]: 1 [0], given: 0 Re: Permutation... [#permalink] 13 Sep 2011, 21:53 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: 183 Location: Singapore Concentration: Finance, General Management GMAT 1: 680 Q46 V37 GMAT 2: 620 Q49 V27 GMAT 3: 700 Q49 V36 WE: Other (Other) Followers: 5 Kudos [?]: 53 [0], given: 43 Re: Permutation... [#permalink] 13 Sep 2011, 23:17 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: 183 Location: Singapore Concentration: Finance, General Management GMAT 1: 680 Q46 V37 GMAT 2: 620 Q49 V27 GMAT 3: 700 Q49 V36 WE: Other (Other) Followers: 5 Kudos [?]: 53 [0], given: 43 Re: Permutation... [#permalink] 13 Sep 2011, 23:21 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: 6219 Location: Pune, India Followers: 1676 Kudos [?]: 9592 [0], given: 197 Re: Permutation... [#permalink] 14 Sep 2011, 21:13 Expert's post 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 My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Permutation... [#permalink]  17 Sep 2011, 05:39
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.

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Re: Permutation... [#permalink]  30 Sep 2011, 02:55
I don't think this type of question will appear on GMAT.... any very good qestion to understand the difference between permutation and combination
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Re: Permutation...   [#permalink] 30 Sep 2011, 02:55
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