Hollywood starlets Ashlee and Britnee each has an entourage

Question

Hollywood starlets Ashlee and Britnee each has an entourage consisting of a number of wannabes. If each starlet has a list of 10 wannabes from which to choose entourage members, which starlet can form the most different entourages?

(1) Britnee’s entourage consists of more wannabes than Ashlee’s entourage does.

(2) Ashlee’s entourage consists of exactly four wannabes.

DmitryFarberMPrep · Accepted Answer

Very clever. This plays on the fact that mathematics is blind to which group is in and which is out. In other words, putting 4 people "in" and 6 "out" is the same as putting 6 "in" and 4 "out." Our intuition is to say that the more members, the more possibilities, but that's not true. In fact, the highest number of possibilities comes at 5 in/5 out, and the number tapers off equally in both directions. There are 252 ways to select a group of 5, 210 ways to select a group of four (or six), etc., all the way down to 1 way to select a group of 10.

(1) INSUFFICIENT:
B could have 5 and A could have 4. B (252) >A (210)
B could have 6 and A could have 5. A (252) >B (210)

(2) INSUFFICIENT:
A's total is 210, but this is useless without info about B.

(1&2) INSUFFICIENT:
A's total is 210.
B could have 5, for a total of 252. B>A
B could have 6, for a total of 210. B=A
B could have 10, for a total of 1. B<A

The correct answer is E.

On a Sentence Correction note, the sentence should read "A&B each have," not "has." The subject is A&B (plural), not each (singular). If this is a real Princeton Review problem, I hope this was a transcription error.

bhandariavi · Answer

Don't you think this problem is little bit difficult to understand especially for non-native English speaker?

RohitKalla · Answer

Yeah it is.. I just assumed entourage for a selected group (word web:The group following and attending to some important person) and wannabes as candidates (word web: An ambitious and aspiring young person). So basically, the no. of people to be selected is different for both the singers. Out of 10 candidates, britnee selects x no. to form her group and ashlee selects y people out of her list of 10 candidates.
Now, statement 2 says- ashlee needs to select 4 out of 10 which means-
there are 10C4 or 210 ways to form ashlee's group. No info about britnee yet.
1st statement says-
Britnee needs to select more people out of 10 than does ashlee. It gives us no clue about the possible no. of groups she can form in comparison to ashlee.
Combining 1st and 2nd-
Britnee needs to select more than 4 people out of 10 to form her group in which case-
No. of ways for britnee ranges from - 10C4 to 10C10 i.e, 252 ways to 1 way.
=>No conclusive answer. Hence, E.
In the process of explaining it to you, I got my doubt cleared

:D Thanks...

gmat7m · Answer

Here's my take:

From 2, If Ashlee has to make entourages of 4 wannabes each and she has 10 to pick from, she could pick 8 people and therefore 2 entourages. But this is not sufficient.
1 seems insufficient on it's own. But if B's entourage is bigger, let's assume 2 numbers: 5 & >5. At 5, she could make two entourages from the 10 people but only 1 entourage at >5. Therefore we don't have a certain number for B.

Therefore,  E ?

rekhabishop · Answer

The question states that "which starlet can form the most different groups".

In that way, the answer should be C.
Because, only 10C1 has the issue of not being greater than 10C(1+n) where n is an integer greater than zero and less than 10.
In all other cases, Britnee's group will have fewer different member teams.

Shobhit7 · Answer

nCk value is maximized when k=n/2. 
If n is odd, then value of n/2+-0.5 is same. Eg. 25C12=25C13.

Similarly, max 10Ck is when k=5.

St.1 is Insufficient because we cannot conclude the higher value of 10Ck even if value of k1 is different from k2.

St.2 is Insufficient as k1= 4.
So, if k2=5, then k2>k1. But, if k2=10, then k2<k1.

Even after combining two statements, we cannot decide which k is leading to higher value of 10Ck.

Hence, Ans E

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hellosanthosh2k2 · Answer

Before we solve this problem, we must note that

10C1 = 10C9, 10C2 = 10C8, 10C3 = 10C7, 10C4 = 10C6

So,

10C1 < 10C2 < 10C3 < 10C4 < 10C5
10C9 < 10C8 < 10C7 < 10C6 < 10C5

Let us attack the statements:

Statement 1: Britnee’s entourage consists of more wannabes than Ashlee’s entourage does.
Say Britnee has 4 wannabes out of 10, and Ashlee has 3 wannabes out of 10 => 10C4 > 10C3, so Britnee has more combinations than Ashless has

Say Britnee has 6 wannabes out of 10, and Ashlee has 5 wannabes out of 10 => 10C6 < 10C4 (from above), so Britnee has less number of combinations than Ashless has

InSuff

Statement2: Ashlee’s entourage consists of exactly four wannabes. - Clearly insufficient, as we don't know about Britnee's

From 1 + 2,
A has 4, but if B has 5 wannabes, then B has more combinations than A has
but if B has 7 wannabes, then effectively 10C7 = 10C3 (<10C4), B has less combinations than A has

Still Insuff -> (E)

bumpbot · Answer

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Hollywood starlets Ashlee and Britnee each has an entourage

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