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Let each different arrangement of all the letters of DELETED

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Director
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Let each different arrangement of all the letters of DELETED [#permalink] New post 19 Oct 2005, 16:12
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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
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 [#permalink] New post 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
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 [#permalink] New post 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?
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Christopher Wilson

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 [#permalink] New post 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
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 [#permalink] New post 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 :cry:

Also, Titelist, do you hear me?!!! Need some 'graphical' help!!
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 [#permalink] New post 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 :cry:
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... 8-)
Director
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 [#permalink] New post 19 Oct 2005, 20:19
6(5!)/3! - This was the missing link!!!

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

:wall

Left side - 'me'
Right side - 'probability'
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 [#permalink] New post 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 :cry:

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!
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 [#permalink] New post 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
  [#permalink] 19 Oct 2005, 20:45
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Let each different arrangement of all the letters of DELETED

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