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# How many words, with or without meaning can be made from the

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How many words, with or without meaning can be made from the [#permalink]  28 Dec 2009, 03:25
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Question Stats:

37% (01:59) correct 62% (00:37) wrong based on 2 sessions
1. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if 4 letters are used at a time?

A. 360
B. 720
C. 240
D. 120
E. 60

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

A. 2419200
B. 25401600
C. 1814400
D. 1926300
E. 1321500

Last edited by zaarathelab on 28 Dec 2009, 05:52, edited 1 time in total.
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Re: Permutation question [#permalink]  28 Dec 2009, 10:30
1.How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if 4 letters are used at a time?

A. 360
B. 720
C. 240
D. 120
E. 60

Choosing 4 letters out of 6 (distinct) letters to form the word = 6C4=15;
Permutations of these 4 letters = 4!=24;

Total # of words possible = 15*24= 360

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

A. 2419200
B. 25401600
C. 1814400
D. 1926300
E. 1321500

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

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Re: Permutation question [#permalink]  28 Dec 2009, 10:53
1. choosing 4 out of 6=6c4=15...
4 can be arranged within themselves=4!... total=360

2. i did it this way..
there are total 12 letters out of which 2 have 4 letters within them and there are 2 T..
fixing P in first posn ,S will come in 6th posn.... the posn can be shifted frm 6 to 12... so 7 places...
p and s can be interchanged... so total way p and s can be arranged=2*7.....
now remaining 10 can be arranged in 10! ways... so total=10!*2*7...
since there are 2 t's.. total ways become 10!*2*7/2!=25401600...B
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Re: Permutation question [#permalink]  28 Dec 2009, 22:46
6c4 - what does this 'c' means?
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Re: Permutation question [#permalink]  28 Dec 2009, 23:27
gmatJP wrote:
6c4 - what does this 'c' means?

c means "combinations"

nCk = C^n_k = \frac{n!}{(n-k)!k!}

Look at this post (Unfortunately, It's not finished yet): math-combinatorics-87345.html
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Re: Permutation question [#permalink]  29 Dec 2009, 02:03
Gr8 explanations guys!

1 A
2 B
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Re: Permutation question [#permalink]  01 Jan 2010, 20:03
sorry, I'm lost on step 4. Why is it 7 rather than 6?

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Re: Permutation question [#permalink]  01 Jan 2010, 20:31
thanks wrote:
sorry, I'm lost on step 4. Why is it 7 rather than 6?

Consider P, S and four letters between them as one unit: {PXXXXS}. 6 more letters are left, so total 7 units: {PXXXXS}, {X}, {X}, {X}, {X}, {X}, {X}. These seven units can be arranged in 7! # of ways.

Hope it's clear.
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Re: Permutation question [#permalink]  11 Apr 2011, 07:03
Bunuel, for question 1, why are we considering 4!?as the problem says, no letter should be repeated.but 4! would mean these 4 letters will repeat.pls help me understand
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Re: Permutation question [#permalink]  11 Apr 2011, 07:25
sdas wrote:
Bunuel, for question 1, why are we considering 4!?as the problem says, no letter should be repeated.but 4! would mean these 4 letters will repeat.pls help me understand

4! doesn't mean that letters are repeated. It means letters are re-arranged.

ABC can be re-arranged in 3!=3*2=6 ways.

ABC
ACB
BAC
BCA
CAB
CBA

If the letters were allowed to repeat: it would be 3*3*3=3^3=27 ways.

1.
6P4 = 6!/2!=6*5*4*3=360
OR
6C4*4!=360

If repetition were allowed, it would be (n^r)
6^4=6*6*6*6=1296 ways
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Re: Permutation question [#permalink]  22 Apr 2011, 17:14
1. 6*5*4*3 = 360

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Re: Permutation question [#permalink]  22 Apr 2011, 17:16
2.

(2* 10*9*7*6 * 7!)/2

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Re: Permutation question [#permalink]  01 May 2011, 00:17
Interesting set indeed.
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Re: Permutation question [#permalink]  27 Dec 2012, 04:35
I used an alternative approach.
P_ _ _ _ S_ _ _ _ _ _. The blank spots can be arranged with the remaining 10 letters in 10! ways. The P and S can together be arranged in 2! ways. Also since it is mentioned that have to be always 4 letters between P and S, hence this arrangement can be stretched to another 6 ways-all together 7 ways. Because of the repition of Ts, divide the entire relation by 2.
Hence,
10!*2*7 / 2 or 10!*7.
Now I shall really appreciate, if anyone helps me in calculating this relation or that of Bunuel's quickly.
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Re: Permutation question [#permalink]  27 Dec 2012, 05:00
Marcab wrote:
I used an alternative approach.
P_ _ _ _ S_ _ _ _ _ _. The blank spots can be arranged with the remaining 10 letters in 10! ways. The P and S can together be arranged in 2! ways. Also since it is mentioned that have to be always 4 letters between P and S, hence this arrangement can be stretched to another 6 ways-all together 7 ways. Because of the repition of Ts, divide the entire relation by 2.
Hence,
10!*2*7 / 2 or 10!*7.
Now I shall really appreciate, if anyone helps me in calculating this relation or that of Bunuel's quickly.

First of all you did everything correct: 10!*7=25,401,600. Next, this is not a GMAT question, because on the exam you won't be asked to calculate 10!*7. If it were a GMAT question, then most likely one of the options would be 10!*7, or we would be able to eliminate other options easily.
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Re: How many words, with or without meaning can be made from the [#permalink]  27 Dec 2012, 20:54
zaarathelab wrote:
1. How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if 4 letters are used at a time?

A. 360
B. 720
C. 240
D. 120
E. 60

Solution 1:
6*5*4*3 = 360

Solution 2:
6!/(2!4!) * 4! = 6!/2! = 6*5*4*3 = 360 Using the Selection/Deselection Formula \frac{N!}{S!D!} then multiplying the selections by 4! to get the arrangement of the 4 letters.

More examples of the Selection/Deselection technique: Combinations: Deselection/Selection
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Re: How many words, with or without meaning can be made from the [#permalink]  27 Dec 2012, 21:04
zaarathelab wrote:
In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and S

A. 2419200
B. 25401600
C. 1814400
D. 1926300
E. 1321500

Solution 1:
1. How many number of ways to position P _ _ _ _ S _ _ _ _ _ _ ? We can have P appear on the first, second, third, to..., seventh of the arrangement of letters. At the same time, we could have P _ _ _ _ S or S _ _ _ _ P.

Thus, 7 * 2!

2. How many ways to arrange 10 remaining letters? 10!/2! We divide by 2! because of 2 Ts in the PERMUTATIONS.

=\frac{7*2!*10!}{2!}=7*10!=25401600

Solution 2:
If you want to understand how permutations work in more detail...

How many number of ways to position P _ _ _ _ S _ _ _ _ _ _ ? We can have P appear on the first, second, third, to..., seventh of the arrangement of letters. At the same time, we could have P _ _ _ _ S or S _ _ _ _ P.

Thus, 7 * 2!

How many number of ways can we select four letters within the P_ _ _ _ S and outside? Getting the number of selections of those selected is always equal to that of those not selected. We use the Selection/Deselection Technique.

10!/4!6!

How many ways can we arrange the four letter within P and S? 4!
How many ways can we arrange the letters outside P and S? 6!
How many duplicate letters just 2 Ts? So we have to divide by 2!.

=\frac{7*2!*10!*4!*6!}{2!*4!*6!}=25401600
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Re: Permutation question [#permalink]  16 Apr 2013, 02:19
Bunuel wrote:
1.How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if 4 letters are used at a time?

A. 360
B. 720
C. 240
D. 120
E. 60

Choosing 4 letters out of 6 (distinct) letters to form the word = 6C4=15;
Permutations of these 4 letters = 4!=24;

Total # of words possible = 15*24= 360

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

A. 2419200
B. 25401600
C. 1814400
D. 1926300
E. 1321500

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

Hey thanks for the solution Bunuel, it is a silly doubt but I did not understand permutations of the 7 units, can you please help me? Why is it not permutations of 6 units?
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Re: Permutation question [#permalink]  16 Apr 2013, 02:44
tox18 wrote:
Bunuel wrote:
1.How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if 4 letters are used at a time?

A. 360
B. 720
C. 240
D. 120
E. 60

Choosing 4 letters out of 6 (distinct) letters to form the word = 6C4=15;
Permutations of these 4 letters = 4!=24;

Total # of words possible = 15*24= 360

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

A. 2419200
B. 25401600
C. 1814400
D. 1926300
E. 1321500

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

Hey thanks for the solution Bunuel, it is a silly doubt but I did not understand permutations of the 7 units, can you please help me? Why is it not permutations of 6 units?

There are 12 letters in "PERMUTATIONS". Four letters between P and S (total of six letters) is one unit: {P(S)XXXXS(P)}, the remaining 6 letters are also one unit each, so total of 7 units.

Hope it's clear.
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Re: Permutation question [#permalink]  16 Apr 2013, 05:06
Bunuel wrote:
tox18 wrote:
Bunuel wrote:
[b]

Hope it's clear.

It is now! Thanks
Re: Permutation question   [#permalink] 16 Apr 2013, 05:06
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