Free GMAT Flashcards available for download!

please help explain ... permutation/comb prob??


 
Post new topic   Reply to topic    Beat The GMAT Forum Index -> GMAT Math -> Problem Solving
View previous topic :: View next topic  
Author Message
karenmeow
Just gettin' started!


Joined: 27 Jun 2008
Posts: 12

Thanks given: 0
Thanked 0 times in 0 posts


PostPosted: Mon Jul 07, 2008 11:26 am    Post subject: please help explain ... permutation/comb prob?? Reply with quote

I copy and pasted from a Kaplan online test (some of the parts are missing in the explanation). In the explanation, it says to divide by 2 because C has to be to the right of D. Where does the 2 come from?? I am lost here. Thanks!

Q: In how many different ways can the letters A, A, B, B. B, C, D, E be arranged if the letter C must be to the right of the letter D?
1,680
2,160
2,520
3,240
3,360

A: We have a total of 8 letters. Some of the letters appear more than once. The requirement of the question stem that the letter C is to be to the right of the letter D, with the letters D and C each appearing once.

If n is a positive integer, the number of different ways to arrange n different objects is n!, where n! is the product of the first n positive integers. Thus, n! = n(n − 1)(n − 2) … (3)(2)(1).

If we had 8 different letters, the number different of ways to arrange the 8 letters would be 8! = (8)(7)(6)(5)(4)(3)(2)(1). Because the letter A appears twice, we must divide 8! by 2!. Because the letter B appears 3 times, we must divide 8! by 3!. So the number of different ways to arrange the letters A, A, B, B, B, C, D, E is 8!/(2!3!). We also have the requirement that the letter C is to the right of the letter D. For any specified locations of the letter C and the letter D, there are the same number of arrangements for the other 6 letters when C is to the right of D and D is to the right of C. So we must divide the number of possible arrangements of all the letters by 2. So the number of different ways to arrange the letters A, A, B, B. B, C, D, E where the letter D is to the right of the letter C is 1/2 (8!/(2!3!)), which is 8!/((2)2!3!). To find the answer to the question, let's find the value. Choice (A) is correct.
Back to top


View user's profile Send private message
pranavc
Really wants to Beat The GMAT!


Joined: 26 Feb 2008
Posts: 160

Thanks given: 9
Thanked 2 times in 2 posts


PostPosted: Mon Jul 07, 2008 1:25 pm    Post subject: Reply with quote

I guess I've gotten quite rusty with my permutations/combinations concepts, but why exactly you divide the total number of possibilities by 2! and 3! for the A's and B's respectively? It would be great if you could fill me in on this bit. Thanks in advance.
Back to top


View user's profile Send private message
evansbd
Rising GMAT Star


Joined: 01 Jul 2008
Posts: 52

Thanks given: 1
Thanked 0 times in 0 posts


Target GMAT Score: 700

PostPosted: Thu Jul 17, 2008 8:01 am    Post subject: Insanity Reply with quote

This question is ridiculous...!
Back to top


View user's profile Send private message
sudhir3127
Moderator


Joined: 07 Jul 2008
Posts: 785

Thanks given: 0
Thanked 59 times in 55 posts

Location: INDIA
Target GMAT Score: 700+

PostPosted: Thu Jul 17, 2008 8:31 am    Post subject: Reply with quote

i think the 2 is because number of way C and D can arrange is 2! which is 2.
Back to top


View user's profile Send private message
CITI29
Rising GMAT Star


Joined: 18 Mar 2008
Posts: 95

Thanks given: 0
Thanked 0 times in 0 posts


PostPosted: Thu Jul 17, 2008 12:54 pm    Post subject: Reply with quote

but if condition is that 'c' has to be to the right of 'd'...then dont we need to treat them as 'fixed' alphabets?.

In such a case solution shld be 7!/ (2!3!).

Pls let me know if my understaning is wrong.
Back to top


View user's profile Send private message
preetha_85
Rising GMAT Star


Joined: 06 Jul 2008
Posts: 95

Thanks given: 2
Thanked 2 times in 2 posts

Location: INDIA
Test Date: Aug 23

PostPosted: Thu Jul 17, 2008 8:24 pm    Post subject: Reply with quote

Hi
The solution goes like this :
U have 8 alphabets nd the condition given is C is to the right of D.
Assume D is in the first pos. then C can g o into any of the other 7 slots nd so can any other alphabet given(i.e 7! ways). But since B is repeated 3 times nd A is repeated 2 twice.. this combination now becomes (7!/(3!*2!)
Now if D is in the 2 nd slot then C can go into only 6 slots (coz of the condition given) then the other alphabets can be arranged in any of the other 6 slots.. so this combination will be (6*(6!/(3!*2!))
Now if D is in 3rd slot then .. (5*(6!/(3!*2!))
nd so on till D is in the last before slot 1*( 6!/(3!*2!))
D in the last slot is not possible.

So the answe will be the sum of all these quantities i.e :
(7!/(3!*2!) + 6!/(3!*2!)) (6+5+4+3+2+1)
i.e 1680
Back to top


View user's profile Send private message
spanlength
Just gettin' started!


Joined: 15 Mar 2008
Posts: 18

Thanks given: 0
Thanked 0 times in 0 posts


PostPosted: Fri Jul 18, 2008 1:58 am    Post subject: Reply with quote

We divide by 2 because C can be either right of D or left of D.So in half of the combination its to the left other half it will to the right.
Back to top


View user's profile Send private message
moliver
Just gettin' started!


Joined: 16 Jul 2008
Posts: 10

Thanks given: 0
Thanked 0 times in 0 posts


PostPosted: Fri Jul 18, 2008 4:45 am    Post subject: Reply with quote

CITI29 wrote:
but if condition is that 'c' has to be to the right of 'd'...then dont we need to treat them as 'fixed' alphabets?.

In such a case solution shld be 7!/ (2!3!).

Pls let me know if my understaning is wrong.


Citi29, C is to the right of D not together.
And I agree with preetha_85
Back to top


View user's profile Send private message
CITI29
Rising GMAT Star


Joined: 18 Mar 2008
Posts: 95

Thanks given: 0
Thanked 0 times in 0 posts


PostPosted: Fri Jul 18, 2008 1:19 pm    Post subject: Reply with quote

ahhhhhh....now I get it...thanks guys!
Back to top


View user's profile Send private message
Display posts from previous:   
Post new topic   Reply to topic    Beat The GMAT Forum Index -> GMAT Math -> Problem Solving All times are GMT - 8 Hours
Page 1 of 1

 
Jump to:  
You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot vote in polls in this forum
You cannot attach files in this forum
You cannot download files in this forum



"GMAT" and other GMAC™ trademarks are registered trademarks of the Graduate Management Admission Council™. The Graduate Management Admission Council™ does not endorse, nor is it affiliated in any way with the owner or any content on this website. The opinions expressed here are solely those of the author or those of the members of this website. Copyright © 2008 BTG Test Prep, LLC. Powered by phpBB © 2001, 2005 phpBB Group.