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# In how many ways can the letters of the word PERMUTATIONS be

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Director
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In how many ways can the letters of the word PERMUTATIONS be [#permalink]

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26 Mar 2008, 17:13
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In how many ways can the letters of the word PERMUTATIONS be arranged if the there are always 4 letters between P and S?

OA is
[Reveal] Spoiler:
25401600
.

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-how-many-ways-can-the-letters-of-the-word-permutations-be-94381.html

Kudos [?]: 340 [0], given: 0

Director
Joined: 10 Sep 2007
Posts: 934

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Re: Permuation Problem [#permalink]

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26 Mar 2008, 17:51
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KUDOS
I figured out the solution however it is not a simple permutation question. This includes both permutation as well as combination.

There are 12 words in letter PERMUTATIONS. Out of which T is repeated twice.

Now first we need to see how many ways we can make word with 4 letter between P and S.
Except P and S there are total of 10 letters, so number of way of selecting them = 10C4 = 210

Also note that question is asking to place exactly 4 words between P and S, but does not tells you if P has to be the first letter of S has to be the first letter. So In all the above combinations, we can rotate the position of P and S.
So total way = 210*2 = 420

The selected 4 letters can be rotated between P and S in = 4! ways

So total ways = 420 * 4!

Consider this 6 letter chunk (P, S, and 4 letter between them) as 1 letter.
Remaining letters are 6. So in total we have 7 letters, which can be arranged in 7! ways.

So total number of ways = 7! * 420 * 4!

Now since letter T was repeated twice, we should divide the above result by 2!.

So Total number of ways = 7! * 420 * 4! / 2! = 25401600

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Manager
Joined: 20 Mar 2010
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Re: Permuation Problem [#permalink]

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24 Mar 2010, 19:51
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Same answer but a slightly different approach.

In the 12 letter word there are 14 different positions (1,6 2,7 3,8 4,9 5,10 6,11 7,12 and reverse) where P and S can be separated by 4 letters. Now the remaining 10 letters can be ordered in 10!/2! ways since T repeats twice.

So total no of ways PERMUTATIONS be arranged so that there are 4 letters between P and S is
14*10!/2! = 25401600

Kudos [?]: 108 [1], given: 1

Senior Manager
Status: Not afraid of failures, disappointments, and falls.
Joined: 20 Jan 2010
Posts: 290

Kudos [?]: 260 [0], given: 260

Concentration: Technology, Entrepreneurship
WE: Operations (Telecommunications)
Re: Permuation Problem [#permalink]

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25 Mar 2010, 03:27
crack700 wrote:
Same answer but a slightly different approach.

In the 12 letter word there are 14 different positions (1,6 2,7 3,8 4,9 5,10 6,11 7,12 and reverse) where P and S can be separated by 4 letters. Now the remaining 10 letters can be ordered in 10!/2! ways since T repeats twice.

So total no of ways PERMUTATIONS be arranged so that there are 4 letters between P and S is
14*10!/2! = 25401600

Sorry! didn't get the red part. Please! can you explain it a bit more?
_________________

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Target=770
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Kudos??

Kudos [?]: 260 [0], given: 260

Intern
Joined: 21 Mar 2010
Posts: 15

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Re: Permuation Problem [#permalink]

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25 Mar 2010, 05:34
2
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AtifS wrote:
crack700 wrote:
Same answer but a slightly different approach.

In the 12 letter word there are 14 different positions (1,6 2,7 3,8 4,9 5,10 6,11 7,12 and reverse) where P and S can be separated by 4 letters. Now the remaining 10 letters can be ordered in 10!/2! ways since T repeats twice.

So total no of ways PERMUTATIONS be arranged so that there are 4 letters between P and S is
14*10!/2! = 25401600

Sorry! didn't get the red part. Please! can you explain it a bit more?

If you start putting in "P" & "S" first you can put them in the following patterns

Total 14 Options
P _ _ _ _ S _ _ _ _ _ _

_ P _ _ _ _ S _ _ _ _ _

_ _ P _ _ _ _ S _ _ _ _

_ _ _ P _ _ _ _ S _ _ _

.
.
.
.
.
_ _ _ _ _ _ P _ _ _ _ S

You can fill the reamining blanks with anyof the letters.

Thanks

Ravi

Kudos [?]: 10 [2], given: 2

Senior Manager
Status: Not afraid of failures, disappointments, and falls.
Joined: 20 Jan 2010
Posts: 290

Kudos [?]: 260 [0], given: 260

Concentration: Technology, Entrepreneurship
WE: Operations (Telecommunications)
Re: Permuation Problem [#permalink]

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25 Mar 2010, 09:32
sudgmat wrote:
AtifS wrote:
crack700 wrote:
Same answer but a slightly different approach.

In the 12 letter word there are 14 different positions (1,6 2,7 3,8 4,9 5,10 6,11 7,12 and reverse) where P and S can be separated by 4 letters. Now the remaining 10 letters can be ordered in 10!/2! ways since T repeats twice.

So total no of ways PERMUTATIONS be arranged so that there are 4 letters between P and S is
14*10!/2! = 25401600

Sorry! didn't get the red part. Please! can you explain it a bit more?

If you start putting in "P" & "S" first you can put them in the following patterns

Total 14 Options
P _ _ _ _ S _ _ _ _ _ _

_ P _ _ _ _ S _ _ _ _ _

_ _ P _ _ _ _ S _ _ _ _

_ _ _ P _ _ _ _ S _ _ _

.
.
.
.
.
_ _ _ _ _ _ P _ _ _ _ S

You can fill the reamining blanks with anyof the letters.

Thanks

Ravi

Thanks! man. kudos
_________________

"I choose to rise after every fall"
Target=770
http://challengemba.blogspot.com
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Kudos [?]: 260 [0], given: 260

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Re: In how many ways can the letters of the word PERMUTATIONS be [#permalink]

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10 Nov 2013, 03:30
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Re: In how many ways can the letters of the word PERMUTATIONS be [#permalink]

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10 Nov 2013, 03:36
abhijit_sen wrote:
In how many ways can the letters of the word PERMUTATIONS be arranged if the there are always 4 letters between P and S?

OA is
[Reveal] Spoiler:
25401600
.

THEORY:

Permutations of $$n$$ things of which $$P_1$$ are alike of one kind, $$P_2$$ are alike of second kind, $$P_3$$ are alike of third kind ... $$P_r$$ are alike of $$r_{th}$$ kind such that: $$P_1+P_2+P_3+..+P_r=n$$ is:

$$\frac{n!}{P_1!*P_2!*P_3!*...*P_r!}$$.

For example number of permutation of the letters of the word "gmatclub" is 8! as there are 8 DISTINCT letters in this word.

Number of permutation of the letters of the word "google" is $$\frac{6!}{2!2!}$$, as there are 6 letters out of which "g" and "o" are represented twice.

Number of permutation of 9 balls out of which 4 are red, 3 green and 2 blue, would be $$\frac{9!}{4!3!2!}$$.

Back to the original questions:

In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S?

There are 12 letters in the word "PERMUTATIONS", out of which T is repeated twice.

1. Choosing 4 letters out of 10 (12-2(P and S)=10) to place between P and S = 10C4 = 210;
2. Permutation of the letters P ans S (PXXXXS or SXXXXP) = 2! =2;
3. Permutation of the 4 letters between P and S = 4! =24;
4. Permutations of the 7 units {P(S)XXXXS(P)}{X}{X}{X}{X}{X}{X} = 7! = 5040;
5. We should divide multiplication of the above 4 numbers by 2! as there is repeated T.

Hence: $$\frac{10C4*2!*4!*7!}{2!}=25,401,600$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-how-many-ways-can-the-letters-of-the-word-permutations-be-94381.html
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Kudos [?]: 129180 [0], given: 12194

Re: In how many ways can the letters of the word PERMUTATIONS be   [#permalink] 10 Nov 2013, 03:36
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# In how many ways can the letters of the word PERMUTATIONS be

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