TestMagic

Swiss · 2004 December 11th, 09:38 PM

On how many ways can the letters of the word "COMPUTER" be arranged?

1) Without any restrictions.
2) M must always occur at the third place.
3) All the vowels are together.
4) All the vowels are never together.
5) Vovels occupy the even posotions.

my soultions..
1)8! 2)7! 3)3!*5! 4)

5)6*3!*5!

does anyone can help me with no.4

Sargataur · 2004 December 14th, 05:42 PM

no.4 answer

= total combinations - all vowles always together
= what u found in 1) - what u found in 3)
= 8! - 6!*3!

ps. actually thy answer for 3) is wrong....considering the 3 vowels as 1 letter ,there r five other letters which r consonants C, M ,P, T, R

CMPTR(AUE) = 6 letters which can be arranged in 6p6 or 6! ways
and A,U,E themselves can be arranged in another 3! ways for a total of 6!*3! ways

HTH

n_adroja · 2005 January 7th, 06:16 PM

can u confirm the answer pls! Specially of ques 3 and 4

thanks
adroja

Quote:

Originally Posted by Swiss

On how many ways can the letters of the word "COMPUTER" be arranged?

1) Without any restrictions.
2) M must always occur at the third place.
3) All the vowels are together.
4) All the vowels are never together.
5) Vovels occupy the even posotions.

my soultions..
1)8! 2)7! 3)3!*5! 4)

5)6*3!*5!

does anyone can help me with no.4

n_adroja · 2005 January 7th, 06:18 PM

Hi Swiss,

can u pls expalin how u solved ques 5!
thanks
adroja

Quote:

Originally Posted by Swiss

On how many ways can the letters of the word "COMPUTER" be arranged?

1) Without any restrictions.
2) M must always occur at the third place.
3) All the vowels are together.
4) All the vowels are never together.
5) Vovels occupy the even posotions.

my soultions..
1)8! 2)7! 3)3!*5! 4)

5)6*3!*5!

does anyone can help me with no.4

GmatGnat · 2005 January 7th, 11:44 PM

My take on question 5 is 5!*3!*2
Option 1: 5*3*4*2*3*1*2*1 = 5!*3! (we run out of vowels for the last even spot)
Option 2: 5*4*3*3*2*2*1*1 = 5!*3! (we start the vowels in the second even spot)
add them up, and you get 5!*3!*2

Swiss, where do you get the 6 from?

katia · 2006 January 25th, 08:32 AM

I think we get only 4 possible cases
2 4 6
4 6 8
2 6 8
2 4 8
thus the final result should be 4*5!*3!

katia

mdlee337 · 2006 January 25th, 06:16 PM

For 5, why is it not 4!*5!?

mac999993 · 2006 February 13th, 08:16 PM

I agree with the proposed solutions for 1) and 2):

1) 8!
2) 7!

3) I became puzzled with the above proposed solution to problem 3) So I conferred with 800Bob, and he agreed with me that 3!*5! is not correct. Thanks Bob. 3!*5! accounts for only 1 variation of this arrangement. There are actually 6:
VVVxxxxx
xVVVxxxx
xxVVVxxx
xxxVVVxx
xxxxVVVx
xxxxxVVV

where VVV are the vowels together, and x are the other consonants.

The solution to 3) is therefore 6*3!*5! = 6*720 = 4320

4) I believe the solution is: Total possible (unrestricted) combinations minus combinations with vowels together (3 above).
8! - 6*3!*5! = 40320-4320=36000

5) There are 4 possible variations to this combination:

xVxVxVxx
xVxVxxxV
xVxxxVxV
xxxVxVxV

So, I believe the solution to 5) is: 4 * 3!*5! = 4*720 = 2880

mac999993 · 2006 February 16th, 11:25 PM

Would someone please validate or correct my solutions above?
NO OG's were provided..

Khilona · 2006 July 23rd, 07:11 AM

First and 2'nd answers correct.
For third, Take Vowels as 1 alphabet. Then total alphabets = 6. So arrangement = 6! But we have to take into consideration that the 3 vowels can be interchanged. So 6! x 3! giving 4320
For 4'th, total possibilities - the previous answer i.e 8! - 6! x 3!
and the last, I solved it as
X_X_X_X_ where x is the place available for the 5 cons. and 4 for vowels
5p4 x 4p3 = 2880

Am with ya Mac

2004 December 11th, 09:38 PM	#1 (permalink)
Swiss I JUST got here. Join Date: Oct 2004 Location: Zurich Posts: 5	Permutations & Combinations On how many ways can the letters of the word "COMPUTER" be arranged? 1) Without any restrictions. 2) M must always occur at the third place. 3) All the vowels are together. 4) All the vowels are never together. 5) Vovels occupy the even posotions. my soultions.. 1)8! 2)7! 3)3!5! 4) 5)63!*5! does anyone can help me with no.4

2004 December 14th, 05:42 PM	#2 (permalink)
Sargataur Within my grasp! Join Date: Nov 2004 Location: Delhi Posts: 367	Re: Permutations & Combinations no.4 answer = total combinations - all vowles always together = what u found in 1) - what u found in 3) = 8! - 6!3! ps. actually thy answer for 3) is wrong....considering the 3 vowels as 1 letter ,there r five other letters which r consonants C, M ,P, T, R CMPTR(AUE) = 6 letters which can be arranged in 6p6 or 6! ways and A,U,E themselves can be arranged in another 3! ways for a total of 6!3! ways HTH _ _ _ _ SIG _ _ _ _ a month to first tame the beast, then lay it down; rip out its legs;then its horns;smash its skull;finally when it's totally helpless, gouge its eyes out and pull out a 750 plus - I'm BLOODTHIRSTY AS HELL

2005 January 7th, 11:44 PM	#5 (permalink)
GmatGnat Within my grasp! Join Date: Nov 2004 Location: Wisconsin, USA Posts: 421	Re: Permutations & Combinations My take on question 5 is 5!3!2 Option 1: 53423121 = 5!3! (we run out of vowels for the last even spot) Option 2: 54332211 = 5!3! (we start the vowels in the second even spot) add them up, and you get 5!3!2 Swiss, where do you get the 6 from?

2006 February 13th, 08:16 PM	#8 (permalink)
mac999993 Eager! Join Date: Jan 2006 Location: Newtown Square, PA Posts: 95	I agree with the proposed solutions for 1) and 2): 1) 8! 2) 7! 3) I became puzzled with the above proposed solution to problem 3) So I conferred with 800Bob, and he agreed with me that 3!5! is not correct. Thanks Bob. 3!5! accounts for only 1 variation of this arrangement. There are actually 6: VVVxxxxx xVVVxxxx xxVVVxxx xxxVVVxx xxxxVVVx xxxxxVVV where VVV are the vowels together, and x are the other consonants. The solution to 3) is therefore 63!5! = 6720 = 4320 4) I believe the solution is: Total possible (unrestricted) combinations minus combinations with vowels together (3 above). 8! - 63!5! = 40320-4320=36000 5) There are 4 possible variations to this combination: xVxVxVxx xVxVxxxV xVxxxVxV xxxVxVxV So, I believe the solution to 5) is: 4 3!5! = 4720 = 2880 Last edited by mac999993 : 2006 February 13th at 09:42 PM.

Thread Tools	Search this Thread
Show Printable Version Email this Page	Search this Thread: Advanced Search
Display Modes
Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

2006 January 25th, 08:32 AM	#6 (permalink)
katia Within my grasp! Join Date: Aug 2005 Posts: 102	I think we get only 4 possible cases 2 4 6 4 6 8 2 6 8 2 4 8 thus the final result should be 45!3! katia

2006 January 25th, 06:16 PM	#7 (permalink)
mdlee337 Eager! Join Date: Oct 2005 Posts: 64	For 5, why is it not 4!*5!?

2006 February 16th, 11:25 PM	#9 (permalink)
mac999993 Eager! Join Date: Jan 2006 Location: Newtown Square, PA Posts: 95	Would someone please validate or correct my solutions above? NO OG's were provided..

2006 July 23rd, 07:11 AM	#10 (permalink)
Khilona Within my grasp! Join Date: Jun 2006 Posts: 126	First and 2'nd answers correct. For third, Take Vowels as 1 alphabet. Then total alphabets = 6. So arrangement = 6! But we have to take into consideration that the 3 vowels can be interchanged. So 6! x 3! giving 4320 For 4'th, total possibilities - the previous answer i.e 8! - 6! x 3! and the last, I solved it as X_X_X_X_ where x is the place available for the 5 cons. and 4 for vowels 5p4 x 4p3 = 2880 Am with ya Mac

TestMagic

Free GMAT, GRE, SAT Test, TOEFL practice tests