permutation n combination

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

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..

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 the future, please provide both the answer choices and the source of the question!

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.

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

You've answered the question:

"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.

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

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

Hi,

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 wrote:

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 the future, please provide both the answer choices and the source of the question!

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

Stuart Kovinsky wrote:

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 the future, please provide both the answer choices and the source of the question!

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 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.

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.

Hi,

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.

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

The simple answer is "because that's not a real equation for solving permutation questions".

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.

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 me explain in another way..

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

Stuart Kovinsky wrote:

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

The simple answer is "because that's not a real equation for solving permutation questions".

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.

duplicate formula works only when you use all N.
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.