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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at
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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, xy, and 5xy are chosen at
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02 Sep 2018, 03:24
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. Exactly that pair and only that pair gives x^2 (by)^2. Please reread the whole discussion.
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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at
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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, xy, and 5xy are chosen at
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21 Oct 2018, 01:51
x^2(by)^2 =(x+by)(xby) only x+y and xy 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, xy, and 5xy are chosen at
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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, xy, and 5xy 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);
\(x^2  (by)^2=(xby)(x+by)\) and only one pair is making such expression: \((x+1*y)(x1*y)\), hence probability is 1/6.
Answer: E.
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Re: If two of the four expressions x+y, x+5y, xy, and 5xy are chosen at
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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, xy, and 5xy are chosen at
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