Forum Home > GMAT > Quantitative > Problem Solving (PS)
Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Dec 1310:00 AM EST
-12:00 PM EST
Have you ever wondered how to score a PERFECT 805 on the GMAT? Meet Julia, a banking professional who used the Target Test Prep course to achieve this incredible feat. Julia's story is nothing short of an inspiration.
Dec 1410:00 AM EST
-12:00 PM EST
Think a 100% GMAT Verbal score is out of your reach? Target Test Prep will make you think again! Our course uses techniques such as topical study and spaced repetition to maximize knowledge retention and make studying simple and fun.
Dec 1411:00 AM IST
-01:00 PM IST
Attend this session to learn how to Deconstruct arguments, Pre-Think Author’s Assumptions, and apply Negation Test.- Dec 15
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Dec 1608:00 AM PST
-11:59 PM PST
GMAT Club 12 Days of Christmas is a 4th Annual GMAT Club Winter Competition based on solving questions. This is the Winter GMAT competition on GMAT Club with an amazing opportunity to win over $40,000 worth of prizes!
25 Jan 2004, 12:39
Kudos
Bookmarks
How many words can be found with the letters of the word PATALIPUTRA without changing the relation order of the vowels and constants?
a.180
b.7200
c.4800
d.3600
e.3200
a.180
b.7200
c.4800
d.3600
e.3200
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block below for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
26 Jan 2004, 13:04
Kudos
Bookmarks
ANS = D.
we need to arrange - C=consonant, V=Vowel
CVCVCVCVCCV
we separate the C's from the V's to find diff. possible arrangements of vowels alone, and consonants alone. then we string them together as above. We get a set of (really) different arrangements of V's and a set of (really) different arrangements of C's and when we string one from each set together we get a (really) new and unique word.
Lets go -
For V's we have to CHOOSE where to put the A's (as they are all the same - we simply choose 3 locations out of the 5 for the A's).
That is 5C3.
then we have two options to place the I and the U must be in the last available place.
Total number of arrangements for V's - 5C3x2x1 = 10x2x1 = 20.
C's - more annoying - we CHOOSE the two locations for the P's 6C2 and then two locations for the T's 4C2 and then place the L in one of two places and the R lands in the last available place.
number of arrangements for C's - 6C2 x 4C2 x2x1 = 15x6x2x1 = 180
total words = 180x20 = 3600.
or D.
Further explanations - why can we separate C's from V's? we will eventually arrange them in the proper order as described above. IF THIS ORDER IS PREDETERMINED we can look at the problem as if it is two separate problems. There is no possibility of interchange between the C's and the V's as they must remain in the same relations one to another!!! we can't create a permutation where C's and V's switch places! So the problem of C's and problem of V's are separate problems.
then there is the counting issue - when we have three A's that are exactly the same (not in diff. colors for example
) then the order of V's - AAIUA is exactly the same as the order AAIUA (see what I mean - I switched the first two A's and you didn't even notice 
).
So we start by deciding where the three places to put the A's are going to be. Each different selection of 3 places will give a different word - but we don't have to switch between the A's because that will leave us with exactly the same word).
If after all that the ans is still wrong
- well, I give up 
we need to arrange - C=consonant, V=Vowel
CVCVCVCVCCV
we separate the C's from the V's to find diff. possible arrangements of vowels alone, and consonants alone. then we string them together as above. We get a set of (really) different arrangements of V's and a set of (really) different arrangements of C's and when we string one from each set together we get a (really) new and unique word.
Lets go -
For V's we have to CHOOSE where to put the A's (as they are all the same - we simply choose 3 locations out of the 5 for the A's).
That is 5C3.
then we have two options to place the I and the U must be in the last available place.
Total number of arrangements for V's - 5C3x2x1 = 10x2x1 = 20.
C's - more annoying - we CHOOSE the two locations for the P's 6C2 and then two locations for the T's 4C2 and then place the L in one of two places and the R lands in the last available place.
number of arrangements for C's - 6C2 x 4C2 x2x1 = 15x6x2x1 = 180
total words = 180x20 = 3600.
or D.
Further explanations - why can we separate C's from V's? we will eventually arrange them in the proper order as described above. IF THIS ORDER IS PREDETERMINED we can look at the problem as if it is two separate problems. There is no possibility of interchange between the C's and the V's as they must remain in the same relations one to another!!! we can't create a permutation where C's and V's switch places! So the problem of C's and problem of V's are separate problems.
then there is the counting issue - when we have three A's that are exactly the same (not in diff. colors for example
) then the order of V's - AAIUA is exactly the same as the order AAIUA (see what I mean - I switched the first two A's and you didn't even notice ).
So we start by deciding where the three places to put the A's are going to be. Each different selection of 3 places will give a different word - but we don't have to switch between the A's because that will leave us with exactly the same word).
If after all that the ans is still wrong
- well, I give up 26 Jan 2004, 14:57
Kudos
Bookmarks
:cool
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
