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?
1) for 13 letters in 13! ways divided by factorial of repeating letters 3 E as 1 letter, 2 A as 1 letter, 2 R as 1 letter, 2 N as 1 letter
13!/3!*2!*2!*2!
If A out of N items are identical, then the number of different permutations of the N items is N!/A!
If a set of N items contains A identical items, B identical items, and C identical items etc.., then the total number of different permutations of N objects is
permutation n combination
This topic has expert replies
Source: Beat The GMAT — Problem Solving |
-
melvinsgmat
- Newbie | Next Rank: 10 Posts
- Posts: 5
- Joined: Sun Jan 17, 2010 5:46 pm
is this solution given above by kstv correct? I think once after we group it into 8 letters, then we need to fix the e and the r and then find the possibility.
Please reply on the same to correct my understanding..
Please reply on the same to correct my understanding..
-
Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
In the future, please provide both the answer choices and the source of the question!Reva wrote: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?
Let's start by removing an E and an R - since they're fixed, they're irrelevant to the number of possibilities; we now have:
MEDITRANEAN
which we can rearrange as:
M D I T R EE AA NN
to better see the unique vs duplicate letters.
And the question becomes:
How many unique 2 letter words can be formed from those letters?
Let's break it down into 2 cases: we start with a unique letter or we start with one of the duplicate letters.
If the first letter is M, D, I, T or R, we have 7 choices for the second letter; that's 5*7 = 35 words.
if the first letter is E, A or N, we have 8 choices for the second letter; that's 3*8 = 24 words.
Accordingly, there are 35 + 24 = 59 possible words we can make.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
-
Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
You've answered the question:kstv wrote: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?
1) for 13 letters in 13! ways divided by factorial of repeating letters 3 E as 1 letter, 2 A as 1 letter, 2 R as 1 letter, 2 N as 1 letter
13!/3!*2!*2!*2!
If A out of N items are identical, then the number of different permutations of the N items is N!/A!
If a set of N items contains A identical items, B identical items, and C identical items etc.., then the total number of different permutations of N objects is
"How many unique words can be formed using all of the letters of the word "MEDITERRANEAN"?
which isn't the question posed here.
The formula only works when you're using all of the objects; here we're only using some of them.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
thanks sir for replying i have a doubt
"Let's break it down into 2 cases: we start with a unique letter or we start with one of the duplicate letters.
If the first letter is M, D, I, T or R, we have 7 choices for the second letter; that's 5*7 = 35 words.
if the first letter is E, A or N, we have 8 choices for the second letter; that's 3*8 = 24 words. "
is there a restriction only with the 2nd letter and with 3rd letter we have 7 option why dont we have 4 options for thr 3rd letter in first case
"Let's break it down into 2 cases: we start with a unique letter or we start with one of the duplicate letters.
If the first letter is M, D, I, T or R, we have 7 choices for the second letter; that's 5*7 = 35 words.
if the first letter is E, A or N, we have 8 choices for the second letter; that's 3*8 = 24 words. "
is there a restriction only with the 2nd letter and with 3rd letter we have 7 option why dont we have 4 options for thr 3rd letter in first case
-
Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
Hi,Reva wrote:thanks sir for replying i have a doubt
"Let's break it down into 2 cases: we start with a unique letter or we start with one of the duplicate letters.
If the first letter is M, D, I, T or R, we have 7 choices for the second letter; that's 5*7 = 35 words.
if the first letter is E, A or N, we have 8 choices for the second letter; that's 3*8 = 24 words. "
is there a restriction only with the 2nd letter and with 3rd letter we have 7 option why dont we have 4 options for thr 3rd letter in first case
we're only selecting 2 letters - the 2 in the middle.
We know that our word is:
E _ _ R
based on the question.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
-
analyst218
- Master | Next Rank: 500 Posts
- Posts: 140
- Joined: Fri Feb 05, 2010 2:43 pm
- Thanked: 3 times
- GMAT Score:720
Stuart Kovinsky wrote:In the future, please provide both the answer choices and the source of the question!Reva wrote: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?
Let's start by removing an E and an R - since they're fixed, they're irrelevant to the number of possibilities; we now have:
MEDITRANEAN
which we can rearrange as:
M D I T R EE AA NN
to better see the unique vs duplicate letters.
And the question becomes:
How many unique 2 letter words can be formed from those letters?
Let's break it down into 2 cases: we start with a unique letter or we start with one of the duplicate letters.
If the first letter is M, D, I, T or R, we have 7 choices for the second letter; that's 5*7 = 35 words.
if the first letter is E, A or N, we have 8 choices for the second letter; that's 3*8 = 24 words.
Accordingly, there are 35 + 24 = 59 possible words we can make.
Stuart, I understand your approach.
Will you explain why 11!/(9!2!2!2!) doesn't work?
E _ _ R so 11C2 with three letter E A N repeating twice?
Might be a stupid question but I need to know where my logic is flawed.
thanks in advance
-
melvinsgmat
- Newbie | Next Rank: 10 Posts
- Posts: 5
- Joined: Sun Jan 17, 2010 5:46 pm
Stuart,I am not able to understand why u took in the second case as 3*8 , which should be also 7 i beleive. can you please clarfiy further.Stuart Kovinsky wrote:In the future, please provide both the answer choices and the source of the question!Reva wrote: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?
Let's start by removing an E and an R - since they're fixed, they're irrelevant to the number of possibilities; we now have:
MEDITRANEAN
which we can rearrange as:
M D I T R EE AA NN
to better see the unique vs duplicate letters.
And the question becomes:
How many unique 2 letter words can be formed from those letters?
Let's break it down into 2 cases: we start with a unique letter or we start with one of the duplicate letters.
If the first letter is M, D, I, T or R, we have 7 choices for the second letter; that's 5*7 = 35 words.
if the first letter is E, A or N, we have 8 choices for the second letter; that's 3*8 = 24 words.
Accordingly, there are 35 + 24 = 59 possible words we can make.
-
Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
Hi,melvinsgmat wrote:
Stuart,I am not able to understand why u took in the second case as 3*8 , which should be also 7 i beleive. can you please clarfiy further.
it's 3*8, because there are 2 of each of those letters.
So, if the first letter is E, A or N, the second letter could be:
M-E-D-I-T-R-A-E-N
in other words, EE, AA and NN are all acceptable choices.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
-
Stuart@KaplanGMAT
- GMAT Instructor
- Posts: 3225
- Joined: Tue Jan 08, 2008 2:40 pm
- Location: Toronto
- Thanked: 1710 times
- Followed by:614 members
- GMAT Score:800
The simple answer is "because that's not a real equation for solving permutation questions".analyst218 wrote: Stuart, I understand your approach.
Will you explain why 11!/(9!2!2!2!) doesn't work?
E _ _ R so 11C2 with three letter E A N repeating twice?
Might be a stupid question but I need to know where my logic is flawed.
thanks in advance
In fact, if you solve your expression you don't even get an integer.
You can use the n!/r!s!t! formula for duplicates only when you're using all n objects; there's no formula that combines the regular permutations formula (n!/(n-k)!) with the duplicate one, which is what you've created.
Many test takers over-rely on formulae for counting questions; fact is, a lot of the time reasoning it out is both more intuitive and faster.
Stuart Kovinsky | Kaplan GMAT Faculty | Toronto
Kaplan Exclusive: The Official Test Day Experience | Ready to Take a Free Practice Test? | Kaplan/Beat the GMAT Member Discount
BTG100 for $100 off a full course
-
Phirozz
- Master | Next Rank: 500 Posts
- Posts: 232
- Joined: Sun Feb 28, 2010 10:47 pm
- Thanked: 10 times
Let me explain in another way..Reva wrote: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?
In MEDITERRANEAN
M - 1
E - 3
D - 1
I - 1
T - 1
R - 2
A - 2
N - 2
Since two letters ie E & R are already fixed, we have left with 11 letters with 2 Es and 1 R
So now we have
M - 1
E - 2
D - 1
I - 1
T - 1
R - 1
A - 2
N - 2
now we have to choose two letters which can be done in two ways
1. both letters are different
here 1st letter can be selected in 8 ways(because 8 different letters are thr) and 2nd letter in 7 ways
so total number of ways is 8*7 = 56
2. Both letters are same
it can be selected in 3 ways is EE, AA and NN
So total number of ways is 56+3 = 59
-
analyst218
- Master | Next Rank: 500 Posts
- Posts: 140
- Joined: Fri Feb 05, 2010 2:43 pm
- Thanked: 3 times
- GMAT Score:720
duplicate formula works only when you use all N.Stuart Kovinsky wrote:The simple answer is "because that's not a real equation for solving permutation questions".analyst218 wrote: Stuart, I understand your approach.
Will you explain why 11!/(9!2!2!2!) doesn't work?
E _ _ R so 11C2 with three letter E A N repeating twice?
Might be a stupid question but I need to know where my logic is flawed.
thanks in advance
In fact, if you solve your expression you don't even get an integer.
You can use the n!/r!s!t! formula for duplicates only when you're using all n objects; there's no formula that combines the regular permutations formula (n!/(n-k)!) with the duplicate one, which is what you've created.
Many test takers over-rely on formulae for counting questions; fact is, a lot of the time reasoning it out is both more intuitive and faster.
there is no formula that combines permutations with duplicate formula.
got it thanks!
I shouldn't rely on formula too much as you said but instead reason it out
Here is another way how to solve it, for me it was more intuitively.
We look unique words looking like this: E _ _ R
We have the following letters: EE AA NN I D M T R
These are 8 different letters.
So not considering the doubles, for the first position we have 8 options, for the second 7, i.e. 8x7=56.
The only options we were missing: EE, AA, and NN: 3
=> 56 + 3 = 59.
We look unique words looking like this: E _ _ R
We have the following letters: EE AA NN I D M T R
These are 8 different letters.
So not considering the doubles, for the first position we have 8 options, for the second 7, i.e. 8x7=56.
The only options we were missing: EE, AA, and NN: 3
=> 56 + 3 = 59.
