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Choir of 5 boys and 6 Girls [#permalink]
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13 Oct 2008, 18:42
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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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Re: Choir of 5 boys and 6 Girls [#permalink]
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13 Oct 2008, 18:57
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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.
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Re: Choir of 5 boys and 6 Girls [#permalink]
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14 Oct 2008, 05:35
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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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Re: Choir of 5 boys and 6 Girls [#permalink]
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14 Oct 2008, 07:16
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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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Re: Choir of 5 boys and 6 Girls [#permalink]
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14 Oct 2008, 07:23
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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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Re: Choir of 5 boys and 6 Girls [#permalink]
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15 Oct 2008, 01:17
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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Re: Choir of 5 boys and 6 Girls [#permalink]
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15 Oct 2008, 01:43
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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Hello,
I have a couple questions. I appreciate your answers! So for M0414, 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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Re: M0414 [#permalink]
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04 Feb 2009, 21:53
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lalalilolila wrote: Hello,
I have a couple questions. I appreciate your answers! So for M0414, 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.
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Re: Choir of 5 boys and 6 Girls [#permalink]
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21 Jul 2010, 12:49
ghettosquad wrote: I would go for B. Its a clear cut E. How did you find B as a suitable option?
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Re: Choir of 5 boys and 6 Girls [#permalink]
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28 Mar 2011, 20:21
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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Re: Choir of 5 boys and 6 Girls [#permalink]
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25 Jul 2011, 16:07
since BBBBB must be together, set BBBBB=B
There are GGGGGG Girl spots so that leaves the below possibility.
_ _ _ _ _ _ _ or 7! in the numerator.
The 6! comes from the total possible number of girl spots (n1)!.
Thus 7! / 6! or 7.



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Re: Choir of 5 boys and 6 Girls [#permalink]
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25 Jul 2011, 16:46
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I used a nonformula 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.
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Re: Choir of 5 boys and 6 Girls [#permalink]
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25 Jul 2011, 17:11
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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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Re: Choir of 5 boys and 6 Girls [#permalink]
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26 Jul 2011, 14:19
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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Re: Choir of 5 boys and 6 Girls [#permalink]
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26 Jul 2011, 23:25
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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Re: Choir of 5 boys and 6 Girls [#permalink]
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27 Jul 2011, 10:02
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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Re: Choir of 5 boys and 6 Girls [#permalink]
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19 Dec 2011, 17:39
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
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Re: Choir of 5 boys and 6 Girls [#permalink]
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27 Jul 2012, 08:48
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So this is how I solved the problem: 5 boys and 6 girls. Because the 5 boys have to be together, I thought of it as one unit. So 1 unit of boys and 6 individual girls= 7 things to arrange.
7!/(1!)(6!) because those 6 girls are all the same things. This equals 7.
Answer is E.



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Re: Choir of 5 boys and 6 Girls [#permalink]
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27 Jul 2012, 11:56
Good one. Learnt from my mistake. I had the arrangement of girls as 6C1. For the boys, instead of considering them as a single way of 5 boys together I considered it as 5C1. Had an option of 11 so went for it. Realized my mistake as soon as the right answer flashed.
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