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How many words, with or without meaning can be made from the [#permalink]
 28 Dec 2009, 03:25
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 Answer: A. 2. In how many ways can the letters of the word PERMUTATIONS be arranged if there are always 4 letters between P and SA. 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!}=25,401,600Answer: B.
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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!
The answers are indeed -
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?
thanks in advance.
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Re: Permutation question [#permalink]
01 Jan 2010, 20:31
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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
Answer is A.
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Re: Permutation question [#permalink]
22 Apr 2011, 17:16
2.
(2* 10*9*7*6 * 7!)/2
Answer is B.
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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!=25401600Solution 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
Answer: A.
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
Answer: B. 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
Answer: A.
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
Answer: B. 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 
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Re: Permutation question
[#permalink]
16 Apr 2013, 05:06
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