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samcot
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Capricorn369
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Ajay369
samcot
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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@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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samcot
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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naveen1003
samcot
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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naveen1003
samcot
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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VeritasPrepKarishma
samcot
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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samcot

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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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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for the ones that calculated, its 24,948,000 aka 13!/(2!)(2!)(2!)(3!)
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