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icandy
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Posted: Mon Oct 13, 2008 6:42 pm |
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Joined: Sat Jul 05, 2008 Posts: 1482 Followers: 26
Kudos (?): 128 (2), given: 1
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The choir consists of 5 boys and 6 girls. In how many ways can the singers be arranged in a row, so that all the boys are together? Do not differentiate between arrangements that are obtained by swapping two boys or two girls. (A) 120 (B) 30 (C) 24 (D) 11 (E) 7 Source: GMAT Club Tests - hardest GMAT questions
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kbulse
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Posted: Tue Oct 14, 2008 5:35 am |
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Joined: Tue Jan 01, 2008 Posts: 250 Schools: Booth, Stern, Haas Followers: 1
Kudos (?): 37 (5), given: 2
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I think we can solve it this way we have 5! ways of aranging boys so that they stay together, then when we put them together we got "goupr of 5 boys" + 6 girls in total seven items to permutate 7! so nomerator will be 5!*7!
now since we cannot differentiate among boys and girls we should discount numerator by 5! (boys) and 6! (girls) (!that's formula for permutation of similar items)
finally we have 5!*7!/5!6!=7 correct me if I am mistaken
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kbulse
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Posted: Tue Oct 14, 2008 7:23 am |
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Joined: Tue Jan 01, 2008 Posts: 250 Schools: Booth, Stern, Haas Followers: 1
Kudos (?): 37 (3), given: 2
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when its written " Do not differentiate between arrangements that are obtained by swapping two boys or two girls." it doesn't mean particularly between two. For example: b1 b2 b3 b4 b5 is the same that b2 b4 b3 b1 b4 b5 or b5 b1 b3 b4 b2 well try this out: http://mdm4u1.wetpaint.com/page/4.3+Per ... l+Elementshope this helps, man
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lalalilolila
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Posted: Wed Feb 04, 2009 1:59 pm |
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Joined: Sat Jan 03, 2009 Posts: 3 Followers: 0
Kudos (?): 1 (1), given: 0
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Hello,
I have a couple questions. I appreciate your answers! So for M04-14, I understand the listings... I don't quite understand the formal notation. 7!/6! . I can't quite wrap my head around the differences in the allotted time..
Does anyone have any strategy on how to improve probability and combination strategy? I have read the wiki guide a couple times ...Are there any good workbooks for them?
There are 7 possibilities:
1. bbbbbgggggg 2. gbbbbbggggg 3. ggbbbbbgggg 4. gggbbbbbggg 5. ggggbbbbbgg 6. gggggbbbbbg 7. ggggggbbbbb
Formally, \frac{7!}{6!} = 7 The correct answer is E.
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vishalgc
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Posted: Mon Oct 13, 2008 6:57 pm |
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Joined: Thu Oct 09, 2008 Posts: 97 Followers: 1
Kudos (?): 4 (0), given: 0
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is it 7?
x gx gx gx gx gx gx
x: group of 5 boys
so there are 7 possible arrangement for gorup of5 boys.
Thanks
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icandy
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Posted: Tue Oct 14, 2008 7:16 am |
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Joined: Sat Jul 05, 2008 Posts: 1482 Followers: 26
Kudos (?): 128 (0), given: 1
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kazakhb wrote: I think we can solve it this way we have 5! ways of aranging boys so that they stay together, then when we put them together we got "goupr of 5 boys" + 6 girls in total seven items to permutate 7! so nomerator will be 5!*7!
now since we cannot differentiate among boys and girls we should discount numerator by 5! (boys) and 6! (girls) (!that's formula for permutation of similar items)
finally we have 5!*7!/5!6!=7 correct me if I am mistaken I got the numerator part. But I did not understand the last part of the Q Do not differentiate between arrangements that are obtained by swapping two boys or two girls.Can some explain the above in layman's terms?
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anuprajan5
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Posted: Wed Oct 15, 2008 1:17 am |
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Joined: Tue Sep 23, 2008 Posts: 15 Followers: 0
Kudos (?): 0 (0), given: 0
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I think the answer is 7!*5! ways of arranging the group+6 girls and 5 ways of interchanging the guys within the group.
Dividing this by the condition 5!*6! will give 7 ways.
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scthakur
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Posted: Wed Oct 15, 2008 1:43 am |
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Joined: Tue Jun 17, 2008 Posts: 1654 Followers: 4
Kudos (?): 94 (0), given: 0
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icandy wrote: Do not differentiate between arrangements that are obtained by swapping two boys or two girls.Can some explain the above in layman's terms?  This means if the first boy is at nth place and second boy is at (n+1)th place, this is the same as if the second boy is at the nth place and first boy is at (n+1)th place.
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GMAT TIGER
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Posted: Wed Feb 04, 2009 9:53 pm |
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Joined: Wed Aug 29, 2007 Posts: 2706 Followers: 28
Kudos (?): 300 (0), given: 19
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lalalilolila wrote: Hello,
I have a couple questions. I appreciate your answers! So for M04-14, I understand the listings... I don't quite understand the formal notation. 7!/6! . I can't quite wrap my head around the differences in the allotted time..
Does anyone have any strategy on how to improve probability and combination strategy? I have read the wiki guide a couple times ...Are there any good workbooks for them?
Here is the question:
The choir consists of 5 boys and 6 girls. In how many ways can the singers be arranged in a row, so that all the boys are together? Do not differentiate between arrangements that are obtained by swapping two boys or two girls.
(C) 2008 GMAT Club - m04#14
* 120 * 30 * 24 * 11 * 7
There are 7 possibilities:
1. bbbbbgggggg 2. gbbbbbggggg 3. ggbbbbbgggg 4. gggbbbbbggg 5. ggggbbbbbgg 6. gggggbbbbbg 7. ggggggbbbbb
Formally, \frac{7!}{6!} = 7 The correct answer is E. assume bbbbb = b so there are 7 places that b can be placed in 7! ways if gggggg, each, were different. and gggggg can be placed in 6! different ways if they each were different. Since they are not, they can be placed in 1! waay. so total ways = 7!/6! = 7.
_________________ Verbal: new-to-the-verbal-forum-please-read-this-first-77546.html Math: new-to-the-math-forum-please-read-this-first-77764.html Gmat: everything-you-need-to-prepare-for-the-gmat-revised-77983.html
GT
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GiorgosAth
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Posted: Wed Jul 21, 2010 5:56 am |
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Joined: Fri Jul 16, 2010 Posts: 4 Followers: 0
Kudos (?): 1 (0), given: 1
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lalalilolila wrote: Hello,
I have a couple questions. I appreciate your answers! So for M04-14, I understand the listings... I don't quite understand the formal notation. 7!/6! . I can't quite wrap my head around the differences in the allotted time..
Does anyone have any strategy on how to improve probability and combination strategy? I have read the wiki guide a couple times ...Are there any good workbooks for them?
There are 7 possibilities:
1. bbbbbgggggg 2. gbbbbbggggg 3. ggbbbbbgggg 4. gggbbbbbggg 5. ggggbbbbbgg 6. gggggbbbbbg 7. ggggggbbbbb
Formally, \frac{7!}{6!} = 7 The correct answer is E. I completely agree.
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gaurav2k101
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Posted: Wed Jul 21, 2010 12:49 pm |
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Joined: Thu Jul 09, 2009 Posts: 57 Location: Bangalore Schools: ISB Followers: 2
Kudos (?): 11 (0), given: 27
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ghettosquad wrote: I would go for B. Its a clear cut E. How did you find B as a suitable option?
_________________ The only way of finding the limits of the possible is by going beyond them into the impossible.
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markbrown525
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Posted: Thu Jul 22, 2010 12:27 am |
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Joined: Thu Feb 25, 2010 Posts: 26 Followers: 0
Kudos (?): 4 (0), given: 0
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I too would love some resources to brush up on my combo/perm q's. Please and Thank you, in advance.
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avellanj2
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Posted: Mon Mar 28, 2011 8:21 pm |
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Joined: Wed Oct 13, 2010 Posts: 2 Followers: 0
Kudos (?): 0 (0), given: 0
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Does the phrase "Do not differentiate between arrangements that are obtained by swapping two boys or two girls" simply mean to not care for the order of boys and girls?? I guess it does mean that.
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snkrhed
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Posted: Mon Jul 25, 2011 4:07 pm |
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Joined: Tue Apr 27, 2010 Posts: 95 Followers: 0
Kudos (?): 4 (0), given: 22
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since BBBBB must be together, set BBBBB=B
There are G-G-G-G-G-G Girl spots so that leaves the below possibility.
_ _ _ _ _ _ _ or 7! in the numerator.
The 6! comes from the total possible number of girl spots (n-1)!.
Thus 7! / 6! or 7.
_________________ GOAL: 7xx
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mourinhogmat1
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Posted: Mon Jul 25, 2011 4:46 pm |
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Joined: Tue Jun 08, 2010 Posts: 296 Followers: 0
Kudos (?): 17 (0), given: 11
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I used a non-formula method. Since position doesn't matter,let the generic arrangement bbbbb. We don't need b1,b2 and so on to differentiate order. So, this reduces the whole group to one collection let's name it A. So, possibilities as follows: Agggggg gAggggg and so on till ggggggA So A has moved 7 times. Posted from my mobile device
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rongali
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Posted: Mon Jul 25, 2011 5:11 pm |
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Joined: Fri Feb 11, 2011 Posts: 117 Followers: 1
Kudos (?): 0 (0), given: 0
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easy one ans is E..bundle all boys as one ..now we have 1 can be placed 7 slots available between the girls
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jw
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Posted: Tue Jul 26, 2011 2:19 pm |
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Status: College student '15 Joined: Mon May 09, 2011 Posts: 19 Location: Florida Followers: 0
Kudos (?): 0 (0), given: 1
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How are you getting the 7!/6!? I do not understand how you see the number of ways to arrange them right off the bat.
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achase
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Posted: Tue Jul 26, 2011 11:25 pm |
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Joined: Wed Jul 06, 2011 Posts: 2 Followers: 0
Kudos (?): 0 (0), given: 0
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jw wrote: How are you getting the 7!/6!? I do not understand how you see the number of ways to arrange them right off the bat. It's because 7!, as described in posts above, is the # of different arrangements of 7 items (6Gs+1group of boys = 6+1=7). Now as a rule of permutations, if items in the group are identical to each other you have to divide the total # of possible arrangements (in this case 7!) by the factorial of the number of identical items. So since our group of 7 consists of 6 identical girls and 1 group of boys, we divide 7! by 6! = 7!/6! = 7 Likewise we need to factor in the different ways we can arrange the group of 5 boys. So we take 5! and divide that by the factorial of identical items within that group of 5, which is 5 (because we have no way of differentiating b1 from b2 or b3 etc.). So that gives us 5!/5! = 1 So 7!/6! * 5!/5! = 7 * 1 = 7 or as other people have written it (7!*5!) / (6!*5!) = 7
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jw
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Posted: Wed Jul 27, 2011 10:02 am |
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Status: College student '15 Joined: Mon May 09, 2011 Posts: 19 Location: Florida Followers: 0
Kudos (?): 0 (0), given: 1
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Ah, so we divide by 6! because it is actually 6!(1!)?
Thanks for the explanation. I guess I will take it for what it is, and I will just remember it for my repertoire.
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Jdam
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Posted: Mon Dec 19, 2011 5:39 pm |
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Joined: Sun Nov 21, 2010 Posts: 148 Followers: 0
Kudos (?): 2 (0), given: 12
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2 ways to solve this problem:
1 is by physically writing each arrangement down... g(5b)ggggg gg(5b)gggg and so on
OR
"Stick" boys together since they are all together. You get 7!/6! = 7
--------------------------- Give Kudos if you like my post!
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