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## Comment on

MISSISSIPPI Rule## Hi Brent,

Can you please help out on this. Since 1st & last digit is fixed, we have to 2 places to be arranged with repetitive elements.

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?

A. 59

B. 11! / (2!*2!*2!)

C. 56

D. 23

E. 11! / (3!*2!*2!*2!)

## Check out Bunuel's great

Check out Bunuel's great solution here: http://gmatclub.com/forum/how-many-different-four-letter-words-can-be-fo...

## If I change question as below

How many different 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?

A. 59

B. 11! / (2!*2!*2!)

C. 56

D. 23

E. 11! / (3!*2!*2!*2!)

will ans. be "B" 11! / (2!*2!*2!)

## That's correct.

That's correct.

Once we place an E in the first space and an R in the last space, we're left with the letters: MDITERANEAN

There are:

- 11 letters in total

- 2 repeated E's

- 2 repeated A's

- 2 repeated N's

So, the total arrangements = 11!/(2! x 2! x 2!)

## Thanks for prompt response.

## Hey! Can you explain how you

On the numerator i got 11*10*9*8*7*6*5*4....and on the numerator i got 4*3*2*1. The numerator and denominator cancel out from 4, 3, 2 and 1, and i was left with 11*10*9*8*7*6*5 and got 1,663,200? Thanks!

-Yvonne

## You have only accounted for

You have only accounted for one 4! in the numerator.

11!/(1!)(4!)(4!)(2!) = (11)(10)(9)(8)(7)(6)(5)(4)(3)(2)(1)/(4)(3)(2)(1)(4)(3)(2)(1)(2)(1)

= (11)(10)(9)(8)(7)(6)(5)/(4)(3)(2)(1)(2)(1)

= (11)(10)(9)(7)(6)(5)/(3)(2)(1)(1)

= (11)(10)(9)(7)(5)

= 34,650

## Hi Brent,

For the word NANNY, I followed the below method:

Stage 1: Finding a place for Letter A

We have 5 options to fill letter A

Stage 2: Finding a place for letter Y

We have 4 options for letter Y

Stage 3: Finding a place for the letters N

As they are all identical and once we place A and Y, the place 3 Ns will take has just 1 option ONLY

Total ways= 5*4*1= 20 ways

Its the same answer but am I doing it correctly?

Thanks!

## That's a perfect approach!

That's a perfect approach!

That works well when there is only 1 set of identical letters/objects.

It becomes more complicated with more than 1 set of identical letters/objects, such as arranging the letters in MISSISSIPPI

Cheers,

Brent

## Many thanks s always!

## So for a word like

I know in your previous lesson you said the GMAT won't ask you to evaluate, but you didn't say the answer here was 11! / 4!4!2!1!, you said the final value. Can you clarify?

## Good question.

Good question.

On the GMAT, it would be highly irregular to be asked to calculate 11!/4!4!2!. Instead, the answer choices would likely be in the form w!/x!y!z!

## Hi Brent,

Could you please explain the below question?

How many words can be formed by taking 4 letters at a time out of the letters of the word MATHEMATICS.

## I've seen that question

I've seen that question several times on the GMAT forums, and I never answer it, because it's far too laborious to be an actual GMAT question.

Mitch (GMATGuruNY - 2nd post) provides a thorough solution here: http://www.beatthegmat.com/combinatorics-t272244.html

That said, I think this question is out of scope for the GMAT.

Cheers,

Brent

## Hi Brent,

I tried to solve both questions in same way, but I think I am doing mistake for DIGIT question, my ans is 48 we can arrange DGT in six ways and then to ensure I is not together we can place them at -D-G-T- . We can place first I in 4 ways and second I in 3 ways so ans will be 6*4*3=72

https://gmatclub.com/forum/the-letters-d-g-i-i-and-t-can-be-used-to-form-5-letter-strings-as-220320.html

https://gmatclub.com/forum/the-letters-d-g-i-i-and-t-can-be-used-to-form-5-letter-strings-as-220320.html

## Question link: https:/

Question link: https://gmatclub.com/forum/the-letters-d-g-i-i-and-t-can-be-used-to-form...

Your approach is perfect, except you have not dealt with some duplication.

Once you start placing the two I's, remember that the I's are identical. So, for example, you might place the first I in space #2 and place the second I in space #3. HOWEVER, this is the SAME as placing the first I in space #3 and placing the second I in space #2.

To account for this duplication, we must divide your answer by 2 to get 36

Cheers,

Brent

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