BREAKING: Target Test Prep releases Brand New 2026 On Demand GMAT prep course

Redeem
TTP OnDemand GMAT Video Course
Featured Course

Target Test Prep GMAT OnDemand

Scott Woodbury-Stewart’s private virtual classroom — 400 hours of master-class video lessons for the GMAT Focus Edition.

  • 715+ score guarantee — highest in the industry (99th percentile)
  • 52 chapters · 1,500+ lessons · 4,000+ practice questions
  • 400 hours of video · 1,500+ instructor-led HD smartboard lessons
  • 300,000+ students accepted to Harvard, Stanford, Wharton, Booth & Sloan & more
  • 24/7 live support + weekly Zoom office hours with GMAT instructors
  • TTP AI Assist — 24/7 AI-powered virtual tutor for instant help
  • 1,200+ flashcards + AI-powered study assistant & daily calendar
  • OnDemand, LiveTeach & GMAT Bootcamp formats available
  • Also: GRE, SAT Math & Executive Assessment courses
  • MBA Admissions Consulting now available
  • 🏆 2025 EdTech Breakthrough Award: Test Prep Solution Provider of the Year
  • 200,000+ students served
  • 5-day free trial — $0 to start, no auto-billing, cancel anytime
★★★★★ 5.0 (559 reviews)
130-pt guarantee
$0 to start then $127/mo
Start Free 5-Day Trial Read 559 Reviews →

P&C

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 91
Joined: Fri Jan 17, 2014 7:34 am
Thanked: 7 times
by parveen110 » Sun Mar 02, 2014 4:34 am
In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

What is wrong in my approach?

# of ways s,s,s,s,p,p,m can be arranged: 7!/4!*2!
now, there are 8 places to insert four I's, that can be done in 8C4 ways.

To sum it up, 8C4*7!/4!*2!

Thank You.
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Sun Mar 02, 2014 6:30 am
parveen110 wrote:In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

When we want to arrange a group of items in which some of the items are identical, we can use something called the MISSISSIPPI rule. It goes like this:
If there are n objects where A of them are alike, another B of them are alike, another C of them are alike, and so on, then the total number of possible arrangements = n!/[(A!)(B!)(C!)....]

Okay, to solve the question, we must recognize the following:
# of "words" where the 4 I's are NOT together = (total # of arrangements) - (# of arrangements where the 4 I's ARE together)

total # of arrangements
Here, we're dealing with the letters M, I, I, I, I, S, S, S, S, P, and P
There are 11 letters in total
There are 4 identical I's
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 11!/[(4!)(4!)(2!)]
= 69,300

# of arrangements where the 4 I's ARE together
To ensure that all 4 I's are together, let's GLUE them together to form just ONE GIANT "LETTER" (IIII). Our letters are now as follows: M, S, S, S, S, P, P, and IIII
There are 8 letters in total
There are 4 identical S's
There are 2 identical P's
So, the total number of possible arrangements = 8!/[(4!)(2!)]
= 840

So, # of "words" where the 4 I's are NOT together = 69,300 - 840 = 68,460

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 16207
Joined: Mon Dec 08, 2008 6:26 pmFollowed by: 1268 members
Location: Vancouver, BC
Thanked: 5254 times
GMAT Score:770

by Brent@GMATPrepNow » Sun Mar 02, 2014 6:33 am
parveen110 wrote: What is wrong in my approach?

# of ways s,s,s,s,p,p,m can be arranged: 7!/4!*2!
now, there are 8 places to insert four I's, that can be done in 8C4 ways.

To sum it up, 8C4*7!/4!*2!
I'm not sure what the part in green means.
What exactly is happening when you have 8 choose 4 (8C4)?
What are the 8 objects and what are you doing with the 4 that you choose?

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
Image
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Senior | Next Rank: 100 Posts
Posts: 91
Joined: Fri Jan 17, 2014 7:34 am
Thanked: 7 times

by parveen110 » Sun Mar 02, 2014 7:43 pm
Brent@GMATPrepNow wrote:
parveen110 wrote: What is wrong in my approach?

# of ways s,s,s,s,p,p,m can be arranged: 7!/4!*2!
now, there are 8 places to insert four I's, that can be done in 8C4 ways.

To sum it up, 8C4*7!/4!*2!
I'm not sure what the part in green means.
What exactly is happening when you have 8 choose 4 (8C4)?
What are the 8 objects and what are you doing with the 4 that you choose?

Cheers,
Brent
Hi Brent,

Thank you for the reply. The text in the green means the following:

I may arrange all the letters except 4 I's in 7!/4!*2!.
Let's say, one of the arrangements may be,
_s_s_s_s_p_p_m_

Now that leaves 8 spaces to arrange 4 I's. Therefore, I may choose any of the four places out of eight in 8C4 ways.

So, total # of arrangements of the word so that all the 4 I's are never together, 8C4*7!/4!*2!.

I hope I have made myself clear. Please clarify the flaw in my approach.

Thanks!
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
GMAT Instructor
Posts: 15539
Joined: Tue May 25, 2010 12:04 pmFollowed by: 1906 members
Location: New York, NY
Thanked: 13060 times
GMAT Score:790

by GMATGuruNY » Mon Mar 03, 2014 6:41 am
parveen110 wrote:
Brent@GMATPrepNow wrote:
parveen110 wrote: What is wrong in my approach?

# of ways s,s,s,s,p,p,m can be arranged: 7!/4!*2!
now, there are 8 places to insert four I's, that can be done in 8C4 ways.

To sum it up, 8C4*7!/4!*2!
I'm not sure what the part in green means.
What exactly is happening when you have 8 choose 4 (8C4)?
What are the 8 objects and what are you doing with the 4 that you choose?

Cheers,
Brent
Hi Brent,

Thank you for the reply. The text in the green means the following:

I may arrange all the letters except 4 I's in 7!/4!*2!.
Let's say, one of the arrangements may be,
_s_s_s_s_p_p_m_

Now that leaves 8 spaces to arrange 4 I's. Therefore, I may choose any of the four places out of eight in 8C4 ways.

So, total # of arrangements of the word so that all the 4 I's are never together, 8C4*7!/4!*2!.

I hope I have made myself clear. Please clarify the flaw in my approach.

Thanks!
The intent of the problem is not crystal clear.
Brent has offered a solution for the following:
How many ways can the letters in MISSISSIPPI be arranged if the four I's cannot occupy four adjacent positions?
Here, it's ok for TWO OR THREE I's to occupy adjacent positions, but it's NOT ok for ALL FOUR I's to occupy adjacent positions.
You have offered a solution for the following:
How many ways can the letters in MISSISSIPPI be arranged if no two I'S can occupy adjacent positions?
Here, it's NOT ok for two or three I's to occupy adjacent positions; there must be at least one consonant between any two successive I's.
Your solution for the latter problem is correct.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1