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How many words, with or without meaning can be made from the [#permalink]
28 Dec 2009, 02:25

5

This post was BOOKMARKED

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

66% (01:57) correct
34% (00:42) wrong based on 57 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, 04:52, edited 1 time in total.

Re: Permutation question [#permalink]
28 Dec 2009, 09:30

1

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

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.

Re: Permutation question [#permalink]
28 Dec 2009, 09:53

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

Re: Permutation question [#permalink]
01 Jan 2010, 19:31

Expert's post

thanks wrote:

sorry, I'm lost on step 4. Why is it 7 rather than 6?

thanks in advance.

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.

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 _________________

"When the going gets tough, the tough gets going!"

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 _________________

Re: Permutation question [#permalink]
27 Dec 2012, 03:35

Expert's post

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. _________________

Re: Permutation question [#permalink]
27 Dec 2012, 04:00

Expert's post

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. _________________

Re: How many words, with or without meaning can be made from the [#permalink]
27 Dec 2012, 19:54

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.

Re: How many words, with or without meaning can be made from the [#permalink]
27 Dec 2012, 20:04

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!.

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?

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.

Re: How many words, with or without meaning can be made from the [#permalink]
04 Jun 2013, 03:48

Bunnel , Need your help on this one. I thought we could have the case like

YYYYYYPXXXXS or PXXXXSYYYYYY

but other combos are also possible like

YYYPXXXXSYYY

In the sense total sum of Ys has to be six but "how many are on P's side and how many on S's side is not fixed " am i missing something fundamental here ?

Re: How many words, with or without meaning can be made from the [#permalink]
05 Jun 2013, 01:48

1

This post received KUDOS

Expert's post

kapilhede17 wrote:

Bunnel , Need your help on this one. I thought we could have the case like

YYYYYYPXXXXS or PXXXXSYYYYYY

but other combos are also possible like

YYYPXXXXSYYY

In the sense total sum of Ys has to be six but "how many are on P's side and how many on S's side is not fixed " am i missing something fundamental here ?

No, your understanding is correct. _________________

Re: Permutation question [#permalink]
10 Jun 2014, 05:56

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.[/quote]

Hi Bunnel,

We have this - - - - - - - P - - - - S , where dashes (-) can be arranged in whatever ways. I took 3 cases : 1. When both T's will be within P and S , 2. - when both will outside of P and S and 3. - when one will be in and other out of Pa nd S.

I get below = ( 2*4!/2!*7! + 2*4!*7!/2! + 2*4!*7!) * 10C4

Re: Permutation question [#permalink]
10 Jun 2014, 06:32

Expert's post

cumulonimbus wrote:

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

Hi Bunnel,

We have this - - - - - - - P - - - - S , where dashes (-) can be arranged in whatever ways. I took 3 cases : 1. When both T's will be within P and S , 2. - when both will outside of P and S and 3. - when one will be in and other out of Pa nd S.

I get below = ( 2*4!/2!*7! + 2*4!*7!/2! + 2*4!*7!) * 10C4

= 2*4!*7!*{1/2+1/2+1) * 10C4 = 2*4!*7!*2 * 10C4

Whats wrong in this calculation ? Please guide

If you put 2 T's between P and S, then multiplying this by 10C4 won't be correct. 10C4 is there the number of ways to choose 4 letter which will be between P and S and if already put 2 T's there then you choose 2 out of 8.

It should be \((2*\frac{4!}{2!}*7!*C^2_8+ 2*4!*\frac{7!}{2!}*C^4_8+ 2*4!*7!*C^3_8)\).

Those are non-GMAT questions. Locking the topic. _________________

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