Let each different arrangement of all the letters of DELETED : PS Archive
Check GMAT Club App Tracker for the Latest School Decision Releases http://gmatclub.com/AppTrack

 It is currently 09 Dec 2016, 03:37

### 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

# Let each different arrangement of all the letters of DELETED

Author Message
Director
Joined: 21 Aug 2005
Posts: 793
Followers: 2

Kudos [?]: 23 [0], given: 0

Let each different arrangement of all the letters of DELETED [#permalink]

### Show Tags

19 Oct 2005, 16:12
00:00

Difficulty:

(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions

### HideShow timer Statistics

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Let each different arrangement of all the letters of DELETED be called a word.
(a) How many words are possible?
(b) In how many of these words will the D's be separated?
VP
Joined: 22 Aug 2005
Posts: 1120
Location: CA
Followers: 1

Kudos [?]: 100 [0], given: 0

### Show Tags

19 Oct 2005, 16:37
(a) total words = 7! / 3! * 2! as there are 3 Es and 2 Ds
= 420

(b) total number of words where both Ds are TOGETHER - count 2 Ds as 1:
number of words = 6!/3! = 120
number of words where Ds are not together = 420 - 120 = 300
Manager
Joined: 04 May 2005
Posts: 133
Location: Chicago
Followers: 1

Kudos [?]: 9 [0], given: 0

### Show Tags

19 Oct 2005, 16:41
A. 7!/(3!2!). 7=total amount of letters, must divide repeated elements. 3=total number of repeating letter E. 2= amount of repeating letter D

B. I will try (the amount above)-(the amount of ways in which D will not be separated). I basically take DD as one letter. However, I do not know how to set it up. Help anyone?
_________________

Christopher Wilson

GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
Followers: 30

Kudos [?]: 348 [0], given: 0

### Show Tags

19 Oct 2005, 18:25
(A) 7!/3!2! (Divide by 3! as we have 3 E's and 2! as we had 2 D's)-> 420

(B) Treat the 2 D's as one entity, then we will have 6!/3! = 120 words where the two D's are together. So # of words where D are aparts = 420-120 = 300
Director
Joined: 21 Aug 2005
Posts: 793
Followers: 2

Kudos [?]: 23 [0], given: 0

### Show Tags

19 Oct 2005, 19:53
OA is 420 and 300.

I am weak in this area. Can someone explain more in detail how we get the 6! in (b)? I am just not able to visualize

Also, Titelist, do you hear me?!!! Need some 'graphical' help!!
SVP
Joined: 05 Apr 2005
Posts: 1731
Followers: 5

Kudos [?]: 72 [0], given: 0

### Show Tags

19 Oct 2005, 20:14
gsr wrote:
OA is 420 and 300.
I am weak in this area. Can someone explain more in detail how we get the 6! in (b)? I am just not able to visualize
Also, Titelist, do you hear me?!!! Need some 'graphical' help!!

total possiblities = 7!/(3! 2!) = 420
total possibilities with D's togather = 6(5!)/3! = 6!/3! = 120
so poss D's seperate = 420 - 120 = 300

in the following ways, we can arrange D's togather so that we get 120:

1st place = DD----- = 2(5!/3!2!) = 5!/3! = 20
2nd place = -DD---- = 5!/3! = 20
3rd place = --DD--- = 5!/3! = 20
4th place = ---DD-- = 5!/3! = 20
5th place = ----DD- = 5!/3! = 20
6th place = -----DD = 5!/3! = 20

total = 120

i am not sure whether it helps or not...
Director
Joined: 21 Aug 2005
Posts: 793
Followers: 2

Kudos [?]: 23 [0], given: 0

### Show Tags

19 Oct 2005, 20:19
6(5!)/3! - This was the missing link!!!

Thanks Himalaya!!
--------------

Left side - 'me'
Right side - 'probability'
GMAT Club Legend
Joined: 07 Jul 2004
Posts: 5062
Location: Singapore
Followers: 30

Kudos [?]: 348 [0], given: 0

### Show Tags

19 Oct 2005, 20:22
gsr wrote:
OA is 420 and 300.

I am weak in this area. Can someone explain more in detail how we get the 6! in (b)? I am just not able to visualize

Also, Titelist, do you hear me?!!! Need some 'graphical' help!!

There are 7 letters, but because we treated the two D's as one entity, it becomes 6 letters now. (We treat D as an entity to calculate number of words where the two D's will be together). Since we now have 6 spaces to fill, so it becomes 6!
SVP
Joined: 05 Apr 2005
Posts: 1731
Followers: 5

Kudos [?]: 72 [0], given: 0

### Show Tags

19 Oct 2005, 20:45
gsr wrote:
6(5!)/3! - This was the missing link!!!
Left side - 'me'
Right side - 'probability'

precisely the above expression is = 2(6)(5!)/(3!2!) = 6!/3! = 120
Display posts from previous: Sort by