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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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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] 21 Oct 2018, 01:51

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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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