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# If two of the four expressions a+b, 3a+b, a-b, and a-3b are

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If two of the four expressions a+b, 3a+b, a-b, and a-3b are [#permalink]  06 May 2012, 09:39
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If two of the four expressions a+b, 3a+b, a-b, and a-3b are chosen at random, what is the probability that their product will be of the form of a^2 + kb^2, where k is a positive integer?

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

Guys,
could you please tell me whether my logic was correct or not?
i solved this task in this way:
4*3=12 --> total number of outcomes
but only two of the expressions will result in desired form -->so the probability of choosing the right pair is 2/12=1/6

but the official explanation is a bit different:

First, determine how many possible combinations there are of the four expressions. Since there are four expressions and we'll choose two of them, we can use the combinations formula:
c=4!/(2!2!)=6
Now determine how many of those combinations fit the requirements. The product must result in an a^2, which rules out 3a+b, since 3a times a will result in 3a^2. The product must also result in a positive k, since the form is +k and k must be positive. That rules out any combination where one of the b terms is positive and the other is negative. The only remaining pair is: (a-b)(a-3b)
That's just one, so the probability of choosing that one is 1/6 , choice (B)
[Reveal] Spoiler: OA
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Re: If two of the four expressions a+b, 3a+b, a-b, and a-3b are [#permalink]  07 May 2012, 09:18
If the question states that k is a POSITIVE integer, then no combination of the equations will give us the form a^2 + kb^2

This is because 5 out of the 6 possible combinations of two terms give you a result with an ab term which does not fit the required form.

The only one that fits the form is (a+b)*(a-b) = a^2 - b^2 in which case k = -1. Therefore for the question to be valid, it should not specifically state that k is a POSITIVE integer.

Now assuming k can be negative, your logic to select the probability is correct as only 1 term out of the possible 6 satisfies the requirement.
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Re: If two of the four expressions a+b, 3a+b, a-b, and a-3b are [#permalink]  08 May 2012, 10:19
the only possible choice is (a+b)(a-b)

so we get 1 out of 6 (2c4=6)
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Re: If two of the four expressions a+b, 3a+b, a-b, and a-3b are   [#permalink] 08 May 2012, 10:19
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