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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] New post 13 Sep 2011, 09:08
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How many words of 4 letters can be formed from the word "MEDITERRANEAN"?
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Re: Permutation... [#permalink] New post 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"?

thanks in advance!!!


Total number of letters = 13, Repetations - 3E, 2R, 2N, 2A
Answer - 13C4/(3!*2!*2!*2!)

Cheers!
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Re: Permutation... [#permalink] New post 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"?

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

Sorry for inconvenience.
Cheers!
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Re: Permutation... [#permalink] New post 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"?

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....
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Re: Permutation... [#permalink] New post 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"?

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.
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Re: Permutation... [#permalink] New post 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"?

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!!!
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Re: Permutation... [#permalink] New post 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"?

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.
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Re: Permutation... [#permalink] New post 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!!!
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Re: Permutation... [#permalink] New post 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.
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Re: Permutation... [#permalink] New post 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!
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Re: Permutation... [#permalink] New post 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.
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Re: Permutation... [#permalink] New post 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.



Loads of Thanks to u,Karishma..:)
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Re: Permutation... [#permalink] New post 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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