I just want to know if my thinking in doing this problem is correct:
If 2 of the 4 expressions x+y, x+5y, x-y, 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 any integer?
4 properties, 2 choices = 4!/2! ways of choosing since order doesn't matter. so 12.
to get the expression x^2-(by)^2 we can only have choices (x+y and x-y) , (x+5y and x-y)
so 2/12 = 1/6 chance.
Probability question
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- cans
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its not correct.
4 properties, 2 choices = 4!/2! ways of choosing since order doesn't matter. so 12.
it will be 4C2. (chose 2 out of 4) = 6.
to get the expression x^2-(by)^2 we can only have choices (x+y and x-y) , (x+5y and x-y)
thus 2/6 = 1/3.
4 properties, 2 choices = 4!/2! ways of choosing since order doesn't matter. so 12.
it will be 4C2. (chose 2 out of 4) = 6.
to get the expression x^2-(by)^2 we can only have choices (x+y and x-y) , (x+5y and x-y)
thus 2/6 = 1/3.
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Cans!!
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- cans
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sorry a mistake on my part.fangtray wrote:... but the correct answer was 1/6...
4C2=6. (total ways)
x+y, x+5y, x-y, 5x-y.. out of these 4 expressions, only (x+y)(x-y) will form the required expression
thus 1/6
If my post helped you- let me know by pushing the thanks button
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Cans!!
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- gmatboost
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(x+5y)(x-y) = x^2 + 4xy - 5y^2...
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