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 →

arrnagement of letters

Expert replies
Post new topic Post Reply
Previous Topic Next Topic
Senior | Next Rank: 100 Posts
Posts: 67
Joined: Thu Jun 26, 2008 8:34 am
Location: india
Thanked: 1 times
by dinaroneo » Sat Dec 10, 2011 6:20 pm
In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S?

The answer is 25401600.

Can anybody explain me how?
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Source: — Problem Solving |
User avatar
GMAT Instructor
Posts: 349
Joined: Wed Sep 28, 2011 3:38 pmFollowed by: 54 members
Location: Austin, TX
Thanked: 236 times
GMAT Score:770

by GmatMathPro » Sat Dec 10, 2011 6:49 pm
First, determine the number of ways to fill out the PXXXXS unit. This unit is 6 letters long: _ _ _ _ _ _. First determine the possibilities for the first and last letters. We have 2 choices for the first letter and 1 for the last letter: 2 _ _ _ _ 1. Now, there are ten remaining letters that could go in the blanks, so 2*10*9*8*7*1 total ways to determine this unit. Now, treat the unit as one and arrange it with the remaining 6 letters which could be done in 7! ways. Multiply these together to get 2*10*9*8*7*1*7!. Finally the letter T appears twice in PERMUTATIONS, so divide all of this by 2 to get 10*9*8*7*7! or [spoiler]7*10!=25401600[/spoiler]
Pete Ackley
GMAT Math Pro
Free Online Tutoring Trial
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
User avatar
Legendary Member
Posts: 588
Joined: Sun Oct 16, 2011 9:42 amFollowed by: 9 members
Location: New Delhi, India
Thanked: 130 times
GMAT Score:720

by rijul007 » Sat Dec 10, 2011 11:52 pm
PERMUTATIONS

P/S _ _ _ _ P/S _ _ _ _ _ _ => 10!/2 * 2
_ P/S _ _ _ _ P/S _ _ _ _ _ => 10!/2 * 2
_ _ P/S _ _ _ _ P/S _ _ _ _ => 10!/2 * 2
_ _ _ P/S _ _ _ _ P/S _ _ _ => 10!/2 * 2
_ _ _ _ P/S _ _ _ _ P/S _ _ => 10!/2 * 2
_ _ _ _ _ P/S _ _ _ _ P/S _ => 10!/2 * 2
_ _ _ _ _ _ P/S _ _ _ _ P/S => 10!/2 * 2

Total no of ways = 7 * 10!/2 * 2 = 25401600
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY

GMAT/MBA Expert

User avatar
GMAT Instructor
Posts: 3835
Joined: Fri Apr 02, 2010 10:00 pmFollowed by: 523 members
Location: Milpitas, CA
Thanked: 1854 times
GMAT Score:770

by Anurag@Gurome » Sun Dec 11, 2011 9:12 pm
dinaroneo wrote:In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S?

The answer is 25401600.

Can anybody explain me how?
No. of letters in the word PERMUTATIONS = 12
If P is placed at 1st blank and S at the 6th place, P _ _ _ _ S _ _ _ _ _ _

Similarly, if P is placed at 2nd blank and S at the 7th place, _ P _ _ _ _ S _ _ _ _ _
This way there are 7 ways of doing so. Also P and S can be interchanged, viz. S at the 1st place and P at the 6th place.
So, no. of ways of placing P and S = 7 * 2 = 14
Corresponding to these 14 ways, the other 10 digits can be placed in 10 places as 10!/2! = 1814400
Therefore, required no. of ways = 14 * 1814400 = 25401600
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
1-800-566-4043 (USA)

Join Our Facebook Groups
GMAT with Gurome
https://www.facebook.com/groups/272466352793633/
Admissions with Gurome
https://www.facebook.com/groups/461459690536574/
Career Advising with Gurome
https://www.facebook.com/groups/360435787349781/
Top
Join the discussion
Log in to postLogin with Googleuse password
QUOTE IN REPLY
Post new topic Post Reply
• Page 1 of 1