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icandy · **Posted:** Mon Oct 13, 2008 6:42 pm


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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

[Reveal] Spoiler: OA

E

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kbulse · **Posted:** Tue Oct 14, 2008 5:35 am

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

kbulse · **Posted:** Tue Oct 14, 2008 7:23 am

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+Elements

hope this helps, man

lalalilolila · **Posted:** Wed Feb 04, 2009 1:59 pm

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.

vishalgc · **Posted:** Mon Oct 13, 2008 6:57 pm

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

icandy · **Posted:** Tue Oct 14, 2008 7:16 am

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?

anuprajan5 · **Posted:** Wed Oct 15, 2008 1:17 am

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.

scthakur · **Posted:** Wed Oct 15, 2008 1:43 am

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.

GMAT TIGER · **Posted:** Wed Feb 04, 2009 9:53 pm

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.

GiorgosAth · **Posted:** Wed Jul 21, 2010 5:56 am

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.

gaurav2k101 · **Posted:** Wed Jul 21, 2010 12:49 pm

ghettosquad wrote:

I would go for B.

Its a clear cut E.

How did you find B as a suitable option?

markbrown525 · **Posted:** Thu Jul 22, 2010 12:27 am

I too would love some resources to brush up on my combo/perm q's. Please and Thank you, in advance.

avellanj2 · **Posted:** Mon Mar 28, 2011 8:21 pm

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.

snkrhed · **Posted:** Mon Jul 25, 2011 4:07 pm

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.

mourinhogmat1 · **Posted:** Mon Jul 25, 2011 4:46 pm

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

rongali · **Posted:** Mon Jul 25, 2011 5:11 pm

easy one ans is E..bundle all boys as one ..now we have 1 can be placed 7 slots available between the girls

jw · **Posted:** Tue Jul 26, 2011 2:19 pm

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.

achase · **Posted:** Tue Jul 26, 2011 11:25 pm

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

jw · **Posted:** Wed Jul 27, 2011 10:02 am

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.

Jdam · **Posted:** Mon Dec 19, 2011 5:39 pm

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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