# Lesson: MISSISSIPPI Rule

## Comment on MISSISSIPPI Rule

### Hi Brent,

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!)

### If I change question as below

If I change question as below...changing four letter words to words

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.

Thanks for prompt response. Your lessons are really helping me

### Hi Brent,

Hi Brent,

Bunuel's solution uses permutations, could you provide an alternate approach using what's covered in the lessons so far in the Counting Module?

Thanks

### Sure thing.

Sure thing.

Here's the question:

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!)

We have: E--R
The remaining letters are: {M, D, I, T, R, EE, AA, NN}

CASE 1: the 2 middle letters are the SAME. The possible words are:
EEER
EAAR
ENNR
There are 3 outcomes

CASE 2: the 2 middle letters are DIFFERENT
Since 2 middle letters are different, then we can ignore the duplicate letters.
So, the remaining letters are {M, D, I, T, R, E, A, N}
Place the 2 middle letters in STAGES

Stage 1: Place 1st middle letter
There are 8 letters to choose from. So, we can complete stage 1 in 8 ways

Stage 2: Place 2nd middle letter
There are now 7 letters to choose from. So, we can complete stage 2 in 7 ways

By the Fundamental Counting Principle, we can complete both stages in (8)(7) ways
So, we can fill in two middle letters in 56 ways

TOTAL NUMBER OF OUTCOMES = 3 + 56 = 59

### Sure thing.

Sure thing.

Here's the question:

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!)

We have: E--R
The remaining letters are: {M, D, I, T, R, EE, AA, NN}

CASE 1: the 2 middle letters are the SAME. The possible words are:
EEER
EAAR
ENNR
There are 3 outcomes

CASE 2: the 2 middle letters are DIFFERENT
Since the 2 middle letters are different, we can ignore the duplicate letters.
So, the remaining letters are {M, D, I, T, R, E, A, N}
Place the 2 middle letters in STAGES

Stage 1: Place 1st middle letter
There are 8 letters to choose from. So, we can complete stage 1 in 8 ways

Stage 2: Place 2nd middle letter
There are now 7 letters to choose from. So, we can complete stage 2 in 7 ways

By the Fundamental Counting Principle, we can complete both stages in (8)(7) ways
So, we can fill in two middle letters in 56 ways

TOTAL NUMBER OF OUTCOMES = 3 + 56 = 59

### Thanks a ton Brent! This is

Thanks a ton Brent! This is really helpful

### Hey! Can you explain how you

Hey! Can you explain how you got 34,650? I got 1,663,200. I had the same # of arrangements at 11!/(1!)(4!)(4!)(2!).

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,

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!

Many thanks s always!

### So for a word like

So for a word like MISSISSIPPI, will one have to do the calculation that leads us to 34650 different arrangements? Seems like a lot of calc

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,

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,

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:/

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

### Dear Brent,

Dear Brent,

I am solving the question 201 on the Official Guide 2018.
Could you explain why I cannot use the MISSISSIPPI rule?

Thanks!

### Question link: https:/

We can use the MISSISSIPPI rule when we want to arrange a group of objects that include IDENTICAL objects.

In question #201, we have 3 males and 3 females, but there is nothing in the question to suggest that the females (or the males) are identical.

Cheers,
Brent

### Hi Brent, didn't quite

Hi Brent, didn't quite understodd your solution for the below question: https://www.beatthegmat.com/counting-question-t287639.html

The question asks for 4 letter word but the solution seems to be of 13 letter word with E and R fixed. Am I missing something here?

### Oops, you're absolutely right

Oops, you're absolutely right!
I have edited my response accordingly.

Cheers,
Brent

### In an earlier lesson you

In an earlier lesson you mentioned that we would not want to calculate 9!. Will there be situations on the GMAT where we will encounter number like 11!? I know that we can simplify by listing out each number then canceling number between the numerator and denominator, but this would still take me a while to calculate.

### If would be highly unlikely

If would be highly unlikely for the GMAT to require you to calculate anything greater than 8!

That said, there have been questions that ask students to determine the number of zeros at the END of a factorial like 20!

Or you might be asked about the nature of a factorial. For example, the following question asks you to determine the greatest integer k for which 2^k is a factor of 20!: https://gmatclub.com/forum/if-n-is-the-product-of-integers-from-1-to-20-...

Cheers,
Brent

### Hi Brad, I can't seem to

Hi Brad, I can't seem to apply MISSIPPI rule for this question:

The letters D, G, I, I, and T can be used to form 5-letter strings such as DIGIT or DGIIT. Using these letters, how many 5-letter strings can be formed in which the two occurrences of the letter I are separated by at least one other letter?

A. 12
B. 18
C. 24
D. 36
E. 48

### Hi g1dm,

Hi g1dm,

If we ignore the restriction, then we can use the MISSISSIPPI to determine the number of ways to arrange the 5 letters.
# arrangements = 5!/2! = 60

Now we must subtract the number arrangements in which the two I's are next to each other.
So, let's "glue" the two I's together to create ONE super-letter.
We now have 4 letters: D, G, T, II
We can arrange 4 objects in 4! ways (= 24 ways)

So, ANSWER = 60 - 24 = 36 = D

ASIDE: Here's another approach: https://gmatclub.com/forum/the-letters-d-g-i-i-and-t-can-be-used-to-form...

Does that help?

Cheers,
Brent

### Hi Brent,

Hi Brent,

In the below question, my answer comes to 90 (6C2*4C2*1). I don't understand why do we need to divide by 3!.

https://gmatclub.com/forum/in-a-sports-club-6-players-are-divided-into-3-teams-of-2-players-how-285654.html

### Question link: https:/

Each outcome in your solution counts each outcome 6 times.
Here's why:

Let A, B, C, D, E, F represent the 6 players.

So, using your approach, we might:
Select A and F for the 1st team
Select D and C for the 2nd team
Select B and E for the 3rd team
So, we get the outcome: AF-DC-BE

Alternatively, using your approach, we might:
Select B and E for the 1st team
Select D and C for the 2nd team
Select A and F for the 3rd team
So, we get the outcome: BE-AF-DC

Notice that the 2 outcomes above are exactly the same. That is, each person has the same teammate.
In fact, your approach also includes: BE-DC-AF, AF-BE-DC, DC-BE-AF and DC-AF-BE

We have counted the same outcome 6 times. For this reason, we must divide 90 by 6 (aka 3!)

ASIDE: The wording of the question is ambiguous (which is why I never listed the question in my Reinforcement Activities box).

Cheers,
Brent

### https://gmatclub.com/forum/a

What does the question by "if there is only one way to present each letter"?

### Question link: https:/

I believe the intent of that proviso is to say that the letters in each set are IDENTICAL.
So, we want to arrange the letters A, A, A, B, B, C, and D (as opposed to the letters A, a, A, B, b, C and d)

Cheers,
Brent

### Hi Brent,

Hi Brent,

In the first example "how many different ways can we arrange the 5 letters in study?" The answer of 5X4X3X2X1 assumes that no letters are to be repeated. In the GMAT, where we are not explicitly told whether letters can be repeated or not repeated, do we assume all letters are to be unique?

Also, in a situation where letters can be repeated in arranging "STUDY", will the answer be something like this: 5x5x5x5x5?

### There's no clear-cut answer

There's no clear-cut answer to your question.

The GMAT may ask us to ARRANGE a set of letters, in which case we must use each of the given letters once.

Sometimes, the question might ask us to use DISTINCT letters or digits, in which case we can't repeat letters or digits.
For example: https://gmatclub.com/forum/a-researcher-plans-to-identify-each-participa...

Other times, the wording will imply that duplicates are allowed.
For example: https://gmatclub.com/forum/a-three-digit-code-for-certain-locks-uses-the...