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Allison and Barbara

by leonswati » Wed Sep 14, 2011 10:45 am
Allison and Barbara are part of an 8-member dance troupe. For the upcoming spring recital, the troupe will be divided into two 4-person ensembles and each ensemble will perform a specialized dance. What fraction of all the possible ensembles that include Allison will also include Barbara?

1/4

3/7

1/2

3/2

6/7









Can someone plz help me solve this....
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by prateek_guy2004 » Wed Sep 14, 2011 1:28 pm
Hi there is a different forum for Problem solving......

This is for DS
Don't look for the incorrect things that you have done rather look for remedies....

https://www.beatthegmat.com/motivation-t90253.html
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by 1947 » Wed Sep 14, 2011 7:37 pm
no clue on this one even after fewe minutes.... :(
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by knight247 » Wed Sep 14, 2011 9:55 pm
The answer is B

Two 4-person ensembles can be made in 2*8C4=140 ways

However we are not interested in the above.

Finding the total number of ways where Allison will always be a part of the ensemble.

2*1C1*7C3=70 Ways .So 70 is the total number of possibilities

Now, the number of ensembles which include Allison that also include Barbara can be found out by computing the number of ensembles that has both of them.

2*2C2*6C2=30 Ways. So 30 is the total number of desired outcomes.

30/70=3/7 Hence B
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by leonswati » Thu Sep 15, 2011 7:02 am
I am still not able to understand it, let me try again...
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by leonswati » Thu Sep 15, 2011 7:09 am
I was wondering why did u divide it by 70 and not by 140... Should read the question correctly... thanks..
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by knight247 » Thu Sep 15, 2011 8:35 am
No problem. This problem was quite a hassle for me when I started a few months ago. But after sheer practice its become kinda easy. Keep practicing as much as possible from all the possible sources especially this site. Cheers
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by thestartupguy » Thu Sep 15, 2011 11:32 am
IMO: B

No. of possible arrangements of two 4 member teams = 8!/(4!)^2 = 70
No. of teams with A and B in the same team = 1 X 1 X 6 X 5 = 30

So, the required answer = 30 / 70 = 3/7
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by mehrasa » Thu Sep 15, 2011 2:47 pm
msr4mba wrote:IMO: B

No. of possible arrangements of two 4 member teams = 8!/(4!)^2 = 70
No. of teams with A and B in the same team = 1 X 1 X 6 X 5 = 30

So, the required answer = 30 / 70 = 3/7
since the Q ask "What fraction of all the possible ensembles that include Allison will also include Barbara?"
we have to find the P(Alison and Barbara in group A OR group B), therefore we have to consider two probability and add to each other
30/70 + 30/70 = 60/70 = 6/7
Ans: E
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by gmatboost » Thu Sep 15, 2011 8:42 pm
The answer that most have given is right, but the numbers are slightly off.

Once we have A in a group, there 7 people left for the other 3 slots. This is 7C3, which is 35, not 70.
7C3 = 7*6*5*4*3*2/(3*2*4*3*2) = 7*6*5/(3*2) = 7*5= 35

Once we have A and B in a group, there 6 people left for the other 2 slots. This is 6C2, which is 15, not 30.
6C2 = 6*5*4*3*2/(2*4*3*2) = 6*5/(2) = 30/2 = 15

So the fraction of the groups that include A that also include B is [spoiler]15/35 which is 3/7.[/spoiler]
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by Deepthi Subbu » Thu Sep 15, 2011 9:29 pm
knight247 wrote:The answer is B

Two 4-person ensembles can be made in 2*8C4=140 ways

However we are not interested in the above.

Finding the total number of ways where Allison will always be a part of the ensemble.

2*1C1*7C3=70 Ways .So 70 is the total number of possibilities

Now, the number of ensembles which include Allison that also include Barbara can be found out by computing the number of ensembles that has both of them.

2*2C2*6C2=30 Ways. So 30 is the total number of desired outcomes.

30/70=3/7 Hence B
I have a question when you multiply by 2 to get the desired outcome - 2*2C2 .An ensemble is a team so ABCD will be similar to BACD . So whats the necessity to find different combinations?
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by kevincanspain » Thu Sep 15, 2011 9:45 pm
3 of the 7 members will be in the same ensemble as Alison, and thus the probability that Barbara is among those 3 is 3/7
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by mehrasa » Thu Sep 15, 2011 10:37 pm
gmatboost wrote:The answer that most have given is right, but the numbers are slightly off.

Once we have A in a group, there 7 people left for the other 3 slots. This is 7C3, which is 35, not 70.
7C3 = 7*6*5*4*3*2/(3*2*4*3*2) = 7*6*5/(3*2) = 7*5= 35

Once we have A and B in a group, there 6 people left for the other 2 slots. This is 6C2, which is 15, not 30.
6C2 = 6*5*4*3*2/(2*4*3*2) = 6*5/(2) = 30/2 = 15

So the fraction of the groups that include A that also include B is [spoiler]15/35 which is 3/7.[/spoiler]
i can not understand why you calculate 7C3, meaning u just considered Alison but not consider Barbara// also, the Q talk about two different group of 4 people...what do you think about the way msr4mba solve the problem??????????
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by gmatboost » Thu Sep 15, 2011 10:50 pm
The two different groups of 4 are irrelevant. The question is:

Of all the possible sets of 4 people that contain A, how many also contain B?

Step 1: How many possible sets of 4 contain A?
Answer: This type of set would look like A123 where the 123 are chosen from among the 7 other people who are not A.
7C3 represents the number of sets of 3 people that can be made out of all the other people besides A.
When we put A into each of these groups, they become 4-person groups that contain A.

Step 2: Of these groups, how many also contain B?
Answer: This type of set would look like AB12 where the 12 are chosen from among the 6 other people who are not A nor B.
6C2 represents the number of sets of 2 people that can be made out of all the other people besides A and B.
When we put A and B into each of these groups, they become 4-person groups that contain A and B.

So, 35 sets of 4 contain A. 15 sets of 4 contain A and B. 15/35 of sets with A also contain B.

Though, I must also say that Kevin's solution is nice.
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by HeintzC2 » Fri Sep 16, 2011 12:18 pm
This is a good solution, but i don't think it is explained very well.
Once Alison is picked, there are 7 members that need to fill 3 spots.
The probability that Barbara is NOT picked is: (6/7)*(5/6)*(4/5), do some cancelling and we get 4/7.
P(Barbara not picked with Alison's Group)= 4/7

This means that P(Barbara picked with Alison's Group) = 1-P(Barbara not picked with Alison's Group)
=1-4/7 = 3/7.

kevincanspain wrote:3 of the 7 members will be in the same ensemble as Alison, and thus the probability that Barbara is among those 3 is 3/7
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