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Team S is to comprise of n debaters chosen from x people? Te

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Manager
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Team S is to comprise of n debaters chosen from x people? Te [#permalink]  09 Oct 2017, 23:13
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Team S is to comprise of n debaters chosen from x people? Team R is comprised of n + 1 debaters chosen from x+1 people.

 Quantity A Quantity B Number of unique team S Number of unique Team R

A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

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[Reveal] Spoiler: OA
Manager
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Re: Team S is to comprise of n debaters chosen from x people? Te [#permalink]  10 Oct 2017, 04:11
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Bunuel wrote:
Team S is to comprise of n debaters chosen from x people? Team R is comprised of n + 1 debaters chosen from x+1 people.

 Quantity A Quantity B Number of unique team S Number of unique Team R

A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given

Kudos for correct solution.

Here the option will be D,

Because we need the values of x and n

If suppose x=2 and n =2

Then
QTYA = 2C2 = 1

QTYB = 3C3 = 1 (since qty B is x+1Cn+1)

Both QTY are equal

Now if we take x=2 and n=1

Then
QTYA = 2C1 = 2

QTYB = 3C2 = 3 (since qty B is x+1Cn+1)

Here QTYB> QTYA.

Hence option is D
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Re: Team S is to comprise of n debaters chosen from x people? Te [#permalink]  15 Oct 2017, 02:07
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A more general solution is:

Quantity A = \frac{x!}{n!(x-n)!}

Quantity B = \frac{(x+1)!}{(n+1)!(x-n)!}

Then, knowing that (x+1)! = (x+1)x! (and the same for n), we can rewrite quantity B as \frac{(x+1)}{(n+1)}\frac{x!}{n!(x-n)!}.

Simplifying we remain with a comparison between 1 and \frac{(x+1)}{(n+1)}. Then, if x = n, quantity B becomes 1 and we have an equality. Otherwise, A is greater.

Thus the answer is D
Re: Team S is to comprise of n debaters chosen from x people? Te   [#permalink] 15 Oct 2017, 02:07
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