Find all School-related info fast with the new School-Specific MBA Forum

 It is currently 04 May 2015, 03:09

### 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

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# The letters of the word PROMISE are arranged so that no two

 Question banks Downloads My Bookmarks Reviews Important topics
Author Message
TAGS:
Director
Joined: 13 Nov 2003
Posts: 793
Location: BULGARIA
Followers: 1

Kudos [?]: 27 [0], given: 0

The letters of the word PROMISE are arranged so that no two [#permalink]  22 Mar 2004, 12:50
The letters of the word PROMISE are arranged so that no two
of the vowels should come together. Find total number of arrangements.

1) 49
2) 1440
3) 4320
4) 1898
Senior Manager
Joined: 02 Mar 2004
Posts: 329
Location: There
Followers: 1

Kudos [?]: 0 [0], given: 0

[#permalink]  22 Mar 2004, 12:58
xPxRxMxSx

4! * 5*4*3 = 1440
Director
Joined: 13 Nov 2003
Posts: 793
Location: BULGARIA
Followers: 1

Kudos [?]: 27 [0], given: 0

[#permalink]  22 Mar 2004, 13:04
Hallo guys, I also got 1440 _P_R_M_S_ or 5C3x4!x3!=1440. there must be a mistake coz the official answer is 4320.Thanx!
Senior Manager
Joined: 02 Mar 2004
Posts: 329
Location: There
Followers: 1

Kudos [?]: 0 [0], given: 0

[#permalink]  22 Mar 2004, 13:08
multiply it with #placing consonants. You will have desired answer
Director
Joined: 13 Nov 2003
Posts: 793
Location: BULGARIA
Followers: 1

Kudos [?]: 27 [0], given: 0

[#permalink]  22 Mar 2004, 13:09
Paul think that you have missed the 4! for the consonants so you answer is also 10x3!x4!. Thanx
GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
Followers: 27

Kudos [?]: 216 [0], given: 0

[#permalink]  22 Mar 2004, 13:11
Let X = vowel
Possible arrangements:

X_X_X_ _
X_X_ _X_
X_X_ _ _X
X_ _X_X_
X_ _X_ _X
X_ _ _X_X
_X_X_X_
_X_X_ _X
_X_ _X_X
_ _X_X_X

In each of above possible ways (10 in total) of placing vowels, you can switch vowels around 3! ways: 3! = 6
Number of ways of arranging consonants: 4! = 24
Since there are 10 ways of arranging vowels: 10(3!*4!) = 1440
Sorry I had to edit my answer because I forgot about consonants
_________________

Best Regards,

Paul

GMAT Club Legend
Joined: 15 Dec 2003
Posts: 4313
Followers: 27

Kudos [?]: 216 [0], given: 0

[#permalink]  22 Mar 2004, 13:12
Wow, quick on correcting mistakes. There was no answer when I started typing and 4 of them popped up by the time I finished! You guys are quick!
_________________

Best Regards,

Paul

Director
Joined: 03 Jul 2003
Posts: 655
Followers: 2

Kudos [?]: 22 [0], given: 0

[#permalink]  22 Mar 2004, 13:15
I'm getting 720.

Total ways to arrange = 7!
there are 3 vowels
ways to pick 2 vowels out of 3 is 3C2 = 3
Ways to arrange the picked two vowels = 2
Ways to arrange 2 vowel combination with other 5 letters = 6!
Total ways to arrange 7 letters with 2 vowels togather = 6 * 6!

Total ways to arrange 7 letter with vowels seperated = 6! = 720
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

[#permalink]  22 Mar 2004, 13:18
Answer should be 4320

No of ways in which vowels are together = no of ways in which exactly two vowels are together + no of ways in which 3 vowels are together.

But if you just choose 2 vowels in 3C2 and consider it as a set , you will have this set, one remaining vowel and 4 other consonants.
If you rearrange these it also covers set containgin 3 vowels together.
so you have 2 * 3C2 * 5! as invalid combinations

No of ways to arrange 7 letters = 7! = 5040

Desired ways = 5040 - 2 * 3C2 * 5! = 4320
Senior Manager
Joined: 02 Mar 2004
Posts: 329
Location: There
Followers: 1

Kudos [?]: 0 [0], given: 0

[#permalink]  22 Mar 2004, 13:28
#no of ways in which 3 vowels are together and #ways 2 vowels are together are not mutually exclusive.
Manager
Joined: 10 Mar 2004
Posts: 64
Location: Dallas,TX
Followers: 1

Kudos [?]: 0 [0], given: 0

[#permalink]  22 Mar 2004, 13:30
anandnk wrote:
Answer should be 4320

No of ways in which vowels are together = no of ways in which exactly two vowels are together + no of ways in which 3 vowels are together.

But if you just choose 2 vowels in 3C2 and consider it as a set , you will have this set, one remaining vowel and 4 other consonants.
If you rearrange these it also covers set containgin 3 vowels together.
so you have 2 * 3C2 * 5! as invalid combinations

No of ways to arrange 7 letters = 7! = 5040

Desired ways = 5040 - 2 * 3C2 * 5! = 4320

Anand,

Just wondering if ur solution takes into account the different ways in which 3 vowels will be together and there placement along with respect to consonants.

cheers
Senior Manager
Joined: 02 Mar 2004
Posts: 329
Location: There
Followers: 1

Kudos [?]: 0 [0], given: 0

[#permalink]  22 Mar 2004, 13:47
I'm getting 720.

Total ways to arrange = 7!
there are 3 vowels
ways to pick 2 vowels out of 3 is 3C2 = 3
Ways to arrange the picked two vowels = 2
Ways to arrange 2 vowel combination with other 5 letters = 6!
Total ways to arrange 7 letters with 2 vowels togather = 6 * 6!

Total ways to arrange 7 letter with vowels seperated = 6! = 720

well, you assumed words such as xxVyVza have no vowles together. For example, x, y or z can be the one Vowel you didn't choose.
Senior Manager
Joined: 11 Nov 2003
Posts: 355
Location: Illinois
Followers: 1

Kudos [?]: 0 [0], given: 0

[#permalink]  22 Mar 2004, 17:18
anandnk wrote:
Answer should be 4320

No of ways in which vowels are together = no of ways in which exactly two vowels are together + no of ways in which 3 vowels are together.

But if you just choose 2 vowels in 3C2 and consider it as a set , you will have this set, one remaining vowel and 4 other consonants.
If you rearrange these it also covers set containgin 3 vowels together.
so you have 2 * 3C2 * 5! as invalid combinations

No of ways to arrange 7 letters = 7! = 5040

Desired ways = 5040 - 2 * 3C2 * 5! = 4320

Your method makes the most sense. Yes 4320 should be the correct answer. Thanks
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

[#permalink]  22 Mar 2004, 18:09
word PROMISE has vowels O,I,E
if you chose two vowels then you will have following combinations

OI - it can also also be arranged as IO
IE - it can also be arranged as EI
OE - it can also be arranged as EO

Consider one combination OI
Now you have PRMSE and OI
Following are some of the combinations that can be obtained
ESMRPOI
ESMRPIO
and
SMRPEIO - > Here all the 3 vowels are together
So just accounting for two vowels at a time should cover everything.

Thus for each vowel pair we have invalid combinations as
2 * 5! -> for OI and IO
2 * 5! -> for IE and EI
2 * 5! -> for EO and OE
this is same as 2 * 3C2 * 5! = 720

Total combinations are 7! = 5040
so valid combinations are 7! - 720 = 4320
SVP
Joined: 30 Oct 2003
Posts: 1797
Location: NewJersey USA
Followers: 5

Kudos [?]: 41 [0], given: 0

[#permalink]  22 Mar 2004, 18:17
Sorry I did a mistake
it should be 2 * 6!

So invalid combinations are 2 * 3C2 * 6! = 4320

Valid combinations are 720 then
Senior Manager
Joined: 02 Mar 2004
Posts: 329
Location: There
Followers: 1

Kudos [?]: 0 [0], given: 0

[#permalink]  22 Mar 2004, 18:22
anandnk wrote:
Sorry I did a mistake
it should be 2 * 6!

So invalid combinations are 2 * 3C2 * 6! = 4320

Valid combinations are 720 then

Now, you can see overlapping permutations.

Quote:
Set A: 2 * 6! -> for OI and IO
Set B: 2 * 6! -> for IE and EI
Set C: 2 * 6! -> for EO and OE

ESMRPOI
ESMRPIO
and
SMRPEIO - > Here all the 3 vowels are together

You will find same permutations in either of the sets A, B, or C--the permutations with 3 vowels together.

For example,
SMRPEIO is counted both in set A(IO/OI) and in set B(EI/IE).
SMRPEOI is counted in set A and in set C and so on.
[#permalink] 22 Mar 2004, 18:22
Similar topics Replies Last post
Similar
Topics:
1 Letter Arrangements in Words 0 24 Sep 2014, 06:06
7 The number of arrangement of letters of the word BANANA in 1 28 Oct 2013, 04:26
Arrange letters of the word 6 20 May 2011, 05:55
In how many ways can the letters of the word ARRANGE be 5 12 Apr 2006, 19:12
In how many ways the letters of the word DOUBLE be arranged 2 04 Sep 2005, 13:05
Display posts from previous: Sort by

# The letters of the word PROMISE are arranged so that no two

 Question banks Downloads My Bookmarks Reviews Important topics

 Powered by phpBB © phpBB Group and phpBB SEO Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.