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Re: Allison and Barbara are part of an 8-member dance troupe [#permalink]
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Assume Allison is in the first ensemble. So there are three more places left and there are 7 members of troupe remaining who could join the first troupe. So the chances of Barbara being one of the remaining three of this ensemble is 3/7.

If Allison is not in the first troupe then she will have to be in the second troupe. So the answer does not change.
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Re: Allison and Barbara are part of an 8-member dance troupe [#permalink]
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PrashantPonde wrote:
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?

A) \(\frac{1}{4}\)

B) \(\frac{3}{7}\)

C) \(\frac{1}{2}\)

D) \(\frac{3}{4}\)

E) \(\frac{6}{7}\)


Let’s first determine the number of ensembles that include Allison. Suppose that Allison has been chosen for one of the ensembles. The three remaining people in the ensemble can be chosen in 7C3 = (7 x 6 x 5)/(3 x 2) = 35 different ways.

Now, let’s determine the number of ensembles that include both Allison and Barbara. Suppose that they are both chosen as members in one of the ensembles. The remaining two people can be chosen in 6C2 = (6 x 5)/(2 x 1) = 15 different ways.

Thus, 15/35 = 3/7 of all the possible ensembles that include Allison also include Barbara.

Answer: B
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Re: Allison and Barbara are part of an 8-member dance troupe [#permalink]
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Total combinations:
C(8,4) = 70

Considering Allison and Barbarba in differente groups:
C(2,1) * C(6,3) = 40
C(6,3) is the number of combinations of 3 members excluding one of them (considering the other one is already in one group)

So, the probability that they would be in different group is:
P = 40/70 = 4/7

Therefore, the probability that are on the same group is:
P = 1 -4/7 = 3/7

Alternative B
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Re: Allison and Barbara are part of an 8-member dance troupe [#permalink]
The sample space will be 8C4. The probability that Allison and Barbara will be in the same ensemble is 6C2, but considering there are 2 ensembles so it needs to be multiplied by 2. Hence we get 3/7.
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Re: Allison and Barbara are part of an 8-member dance troupe [#permalink]
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PrashantPonde wrote:
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?

A) \(\frac{1}{4}\)

B) \(\frac{3}{7}\)

C) \(\frac{1}{2}\)

D) \(\frac{3}{4}\)

E) \(\frac{6}{7}\)


Number of ways to choose 4 members from 8 members such that Allison is always included = 7C3 = 35
(A _ _ _) Total left: 7, To pick: 3
Number of ways to choose 4 members from 8 members such that Allison and Barbara are always together = 6C2 = 15
(A B _ _) Total left: 6, To pick: 2
Therefore fraction of all the possible ensembles that include Allison and will also include Barbara = 15/35 = 3/7
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Re: Allison and Barbara are part of an 8-member dance troupe [#permalink]
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Re: Allison and Barbara are part of an 8-member dance troupe [#permalink]
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