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

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

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Here a couple of problems I encountered. Let me know your views on them. They may not necessarily be of GMAT format. Please share your views, nevertheless.

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

2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

Cheers
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Re: Interesting problems of Permutations and Combinations [#permalink]

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New post 18 May 2010, 01:13
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chandrun wrote:
Here a couple of problems I encountered. Let me know your views on them. They may not necessarily be of GMAT format. Please share your views, nevertheless.

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

2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

Cheers


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:

1. 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: \(\frac{10C4*2!*4!*7!}{2!}=25,401,600\)

2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

There are 11 letters in the word "MISSISSIPPI ", out of which: M=1, I=4, S=4, P=2.

Total # of permutations is \(\frac{11!}{4!4!2!}\);
# of permutations with 4 I's together is \(\frac{8!}{4!2!}\). Consider 4 I's as one unit: {M}{S}{S}{S}{S}{P}{P}{IIII} - total 8 units, out of which {M}=1, {S}=4, {P}=2, {IIII}=1.

So # of permutations with 4 I's not come together is: \(\frac{11!}{4!4!2!}-\frac{8!}{4!2!}\).

Hope it helps.
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Re: Interesting problems of Permutations and Combinations [#permalink]

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New post 18 May 2010, 07:09
Thank you bunuel, Doing great job dude...
Hats off!!!!
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Re: Interesting problems of Permutations and Combinations [#permalink]

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New post 19 May 2010, 08:13
Thank you very much. That is an awesome piece of explanation.
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Re: Interesting problems of Permutations and Combinations [#permalink]

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New post 31 Jan 2011, 17:31
Why 7!? There are only six letters to arrange...
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Re: Interesting problems of Permutations and Combinations [#permalink]

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New post 31 Jan 2011, 17:45
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mariyea wrote:
Why 7!? There are only six letters to arrange...


Are you talking about Q1, 4th point? Can you please be more specific when asking questions?

4. Permutations of the 7 units {P(S)XXXXS(P)}{X}{X}{X}{X}{X}{X} = 7! = 5040;

There are 7 units: {PXXXXS}, {X}, {X}, {X}, {X}, {X}, {X} --> 6 letters in unit 1, plus 6 units with one letter in each = total of 12 letters.
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Re: Interesting problems of Permutations and Combinations [#permalink]

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New post 01 Feb 2011, 07:07
Bunuel wrote:
mariyea wrote:
Why 7!? There are only six letters to arrange...


Are you talking about Q1, 4th point? Can you please be more specific when asking questions?

4. Permutations of the 7 units {P(S)XXXXS(P)}{X}{X}{X}{X}{X}{X} = 7! = 5040;

There are 7 units: {PXXXXS}, {X}, {X}, {X}, {X}, {X}, {X} --> 6 letters in unit 1, plus 6 units with one letter in each = total of 12 letters.


Yes I am referring to the first q. Sorry about that.

But the q asks for four letters to be placed b/n P and S... and the ways in which the four letters can be arranged is expressed by 4!
How can there be six units that have one letter each?
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Re: Interesting problems of Permutations and Combinations [#permalink]

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New post 01 Feb 2011, 07:26
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mariyea wrote:
Yes I am referring to the first q. Sorry about that.

But the q asks for four letters to be placed b/n P and S... and the ways in which the four letters can be arranged is expressed by 4!
How can there be six units that have one letter each?


It seems that you don't understand what the question is asking. It does not ask about the ways 4 lettter can be arranged between P and S.

The question is: in how many ways can the word PERMUTATIONS be arranged SO THAT in each arrangement there are always 4 letters between P and S.

This should be calculated in several steps. Step 4 is dealing with arrangement of 7 units: P and S with 4 letters between them (as required) is one unit {PXXXXS} so we used 6 letters, EACH of the rest 6 letters is a separate unit itself so we have total of 7 units: {PXXXXS}, {X}, {X}, {X}, {X}, {X}, {X} --> 7!.
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Re: Interesting problems of Permutations and Combinations [#permalink]

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New post 01 Feb 2011, 07:39
Bunuel wrote:
mariyea wrote:
Yes I am referring to the first q. Sorry about that.

But the q asks for four letters to be placed b/n P and S... and the ways in which the four letters can be arranged is expressed by 4!
How can there be six units that have one letter each?


It seems that you don't understand what the question is asking. It does not ask about the ways 4 lettter can be arranged between P and S.

The question is: in how many ways can the word PERMUTATIONS be arranged SO THAT in each arrangement there are always 4 letters between P and S.

This should be calculated in several steps. Step 4 is dealing with arrangement of 7 units: P and S with 4 letters between them (as required) is one unit {PXXXXS} so we used 6 letters, EACH of the rest 6 letters is a separate unit itself so we have total of 7 units: {PXXXXS}, {X}, {X}, {X}, {X}, {X}, {X} --> 7!.


Bunuel, Bunuel! Thank you so much! I get it now, you're the man!
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Re: Interesting problems of Permutations and Combinations [#permalink]

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New post 03 Feb 2011, 03:42
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Bunuel wrote:
chandrun wrote:
Here a couple of problems I encountered. Let me know your views on them. They may not necessarily be of GMAT format. Please share your views, nevertheless.

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

2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

Cheers


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:

1. 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: \(\frac{10C4*2!*4!*7!}{2!}=25,401,600\)

2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

There are 11 letters in the word "MISSISSIPPI ", out of which: M=1, I=4, S=4, P=2.

Total # of permutations is \(\frac{11!}{4!4!2!}\);
# of permutations with 4 I's together is \(\frac{8!}{4!2!}\). Consider 4 I's as one unit: {M}{S}{S}{S}{S}{P}{P}{IIII} - total 8 units, out of which {M}=1, {S}=4, {P}=2, {IIII}=1.

So # of permutations with 4 I's not come together is: \(\frac{11!}{4!4!2!}-\frac{8!}{4!2!}\).

Hope it helps.



What about my approach to PERMUTATIONS?

The word permutations consists of 12 letters.
You can choose P and S in 7*2 ways so there are always 4 numbers between them
P on first, S on fifth ........ P on seventh S on twelwth + reversely (SP)

You are left with 10 letters, 2 of which are the same (TT)

So the complete formula is:

\(7*2*\frac{10!}{2!}=7*10!\)

It makes the same result, but I think is a bit quicker.
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Re: Interesting problems of Permutations and Combinations [#permalink]

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New post 03 Feb 2011, 03:48
craky wrote:
Bunuel wrote:
chandrun wrote:
Here a couple of problems I encountered. Let me know your views on them. They may not necessarily be of GMAT format. Please share your views, nevertheless.

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

2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

Cheers


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:

1. 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: \(\frac{10C4*2!*4!*7!}{2!}=25,401,600\)

2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

There are 11 letters in the word "MISSISSIPPI ", out of which: M=1, I=4, S=4, P=2.

Total # of permutations is \(\frac{11!}{4!4!2!}\);
# of permutations with 4 I's together is \(\frac{8!}{4!2!}\). Consider 4 I's as one unit: {M}{S}{S}{S}{S}{P}{P}{IIII} - total 8 units, out of which {M}=1, {S}=4, {P}=2, {IIII}=1.

So # of permutations with 4 I's not come together is: \(\frac{11!}{4!4!2!}-\frac{8!}{4!2!}\).

Hope it helps.



What about my approach to PERMUTATIONS?

The word permutations consists of 12 letters.
You can choose P and S in 7*2 ways so there are always 4 numbers between them
P on first, S on fifth ........ P on seventh S on twelwth + reversely (SP)

You are left with 10 letters, 2 of which are the same (TT)

So the complete formula is:

\(7*2*\frac{10!}{2!}=7*10!\)

It makes the same result, but I think is a bit quicker.


This approach is also correct.
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Re: Interesting problems of Permutations and Combinations [#permalink]

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New post 04 Feb 2011, 06:41
1. In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S?


PERMUTATIONS:
Total length of the word = 12
Repetitions:
T:2. Rest all letters:1

P,S can occupy following indices in the word
{1,6},{2,7},{3,8},{4,9},{5,10},{6,11},{7,12}

In each of the above positions that P and S occupy, the remaining 10 letters can be arranged in
\(\frac{10!}{2!}\) ways

Likewise; S,P can occupy following indices:
{1,6},{2,7},{3,8},{4,9},{5,10},{6,11},{7,12}

In each of the above positions that S and P occupy, the remaining 10 letters can be arranged in
\(\frac{10!}{2!}\) ways

So, total number of arrangements:

\(\frac{7*2*10!}{2!}\)


2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

Word MISSISSIPPI
Length: 11
Repetitions:
I:4
S:4
P:2
M:1

Total number of arrangement possible = \(\frac{11!}{4!4!2!}\) ------------Ist

Now, let's conjoin four I's and treat it as a unique character, say #. We have conjoined all fours in order to symbolize that all I's are adhered together.

So, Now MISSISSIPPI becomes M#SSSSPP
Word: M#SSSSPP
Length: 8
Repetitions:
S:4
P:2
M:1
#:1

The number of ways in which I's come together is:
\(\frac{8!}{2!4!}\) --------- 2nd

Thus, the number of arrangements in which I's not come together will be the difference between 2nd and Ist

\(\frac{11!}{4!4!2!}-\frac{8!}{2!4!}\)
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Re: Interesting problems of Permutations and Combinations [#permalink]

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New post 05 Feb 2011, 04:59
geeeesh bunuel... amazing explanation.
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Re: In how many ways can the letters of the word PERMUTATIONS be [#permalink]

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New post 28 Aug 2014, 22:19
In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S?
I solved this way:
P_ _ _ _ S _ _ _ _ _ _

10! [ arrange remaining 10 letters]
divided by 2! [ because of 2 T]
multiplied by 2! [ swap P and S]
multiplied by 7 [ P and S can move 6 places ahead]

= 7.10!
Please tell if correct approach?
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Re: In how many ways can the letters of the word PERMUTATIONS be [#permalink]

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New post 05 Jun 2016, 03:13
1
Bunuel wrote:
chandrun wrote:
Here a couple of problems I encountered. Let me know your views on them. They may not necessarily be of GMAT format. Please share your views, nevertheless.

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

2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

Cheers


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:

1. 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;

Hi,

As you have chosen 4 letters out of 10 letters which include two TT's, would it not lead to duplication of the selections?
As in the case of "ENTRANCE", if we had to choose the number of selections, we would go as follows -

We have 8 letters from which 6 are unique.

Possible scenarios for 4 letter selection are:
A. All letters are different;
B. 2 N-s and other letters are different;
C. 2 E-s and other letters are different;
D. 2 N-s, 2 E-s.

Let's count them separately:
A. All letters are different, means that basically we are choosing 4 letters form 6 distinct letters: 6C4=15;
B. 2 N-s and other letters are different: 2C2(2 N-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;
C. 2 E-s and other letters are different: 2C2(2 E-s out of 2)*5C2(other 2 letters from distinct 5 letters left)=10;;
D. 2 N-s, 2 E-s: 2C2*2C2=1.

15+10+10+1=36
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Re: In how many ways can the letters of the word PERMUTATIONS be [#permalink]

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New post 15 Sep 2016, 02:59
is this a gmat type question?
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Re: In how many ways can the letters of the word PERMUTATIONS be [#permalink]

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New post 06 May 2017, 05:46
I am too a little confused. If we have two T's then number of combination should go down while selecting 4 numbers out of 10.
E.g. AABC, If we have to choose 2 letters from these 4 then we will have following options:
AB, BC, AC, AA = 4
But if we apply combination formula 4C2 then answer will be 6.

A little confused, if anyone can guide.
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Re: In how many ways can the letters of the word PERMUTATIONS be [#permalink]

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New post 09 May 2018, 02:12
Bunuel
Sorry to bump in an old topic but my doubt is concerned with the topic in discussion.

Q. Possible arrangements for the word REVIEW if one E can't be next to the other.

My approach:

_R_V_I_W_ .Since both E acn't be together so they can be arranged in the 5 blank spaces.

5C2 * 4!(5C2 for selecting 2 out of 5 places for E & rest can be arranged in 4! ways.)

Ans with this approach 240 which is correct in this case.

Now the question discussed above:

2. In how many of the distinct permutations of the letters in the word MISSISSIPPI do the 4 I's not come together?

Going as per the above approach it comes

8C4 * (7!/4!2!) (8C4 for filling 4 I's in 8 blank spaces and arranging rest 7 alphabets in 7!/4!2!)
But in this case answer doesn't match.

Please if you could explain the flaw in my approach will be very helpful.
Cheers.
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Re: In how many ways can the letters of the word PERMUTATIONS be   [#permalink] 09 May 2018, 02:12
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