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If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at

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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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New post 02 Sep 2018, 03:07
Why the pair (x + y) and (x - y) is not considered? It would yield x² - y² and there is an integer (1) multiplying y.
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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New post 02 Sep 2018, 03:24
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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New post 02 Sep 2018, 12:29
thiagod wrote:
Why the pair (x + y) and (x - y) is not considered? It would yield x² - y² and there is an integer (1) multiplying y.


That is the pair. That's the only pair that yields the correct product.

And there are 4C2 ways to pick a pair. 4C2 = \(\frac{4!}{2!2!}\) = 6.

So the there's only 1 pair that satisfies out of the 6 possible pairs that could be chosen. \(\frac{1}{6}\)

Hope this helps!
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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New post 21 Oct 2018, 01:51
x^2-(by)^2
=(x+by)(x-by)
only x+y and x-y can be the pair
All occasions are "selection of 2 out of 4 = 4C2= 6
Probability is 1/6
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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New post 15 Jul 2019, 07:34
total expressions possible 4c2 ; 6
so x^2 -(by)^2 ; 1/6
IMO E

Bunuel wrote:
If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random, what is the probability that their product will be of the form of x^2 -(by)^2, where b is an integer?

A. 1/2
B. 1/3
C. 1/4
D. 1/5
E. 1/6

Four expressions can be paired in \(C^2_4=6\) ways (choosing 2 out of 4);

\(x^2 - (by)^2=(x-by)(x+by)\) and only one pair is making such expression: \((x+1*y)(x-1*y)\), hence probability is 1/6.

Answer: E.
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Re: If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at  [#permalink]

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New post 17 Jul 2019, 15:24
Made the silly mistake of including (x+5y) in the selection range.
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Re: If two of the four expressions x + y, x + 5y, x – y,...  [#permalink]

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