If two of the four expressions x + y , x + 3y , x - y , and 3x - 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? Select
A) 1/2  B) 1/6  C) 1/5  D) 1/3  E) 1/4
Total number of expressions if 2 are multiplied = 4C2 = 4!/2!*2! = 6   now we have to find the expressions that are in form of x^2 - (by)^2   one is (x+y)(x-y) = x^2 - (1y)^2   If you see any other combination we are always going to have one term of xy since 3x and 3y are there, so there can be only one type of combination like this.   therefore, probablity is 1/6   Answer : 1/6
<result>B</result>
If two of the four expressions x + y , x + 3y , x - y , and 3x - 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?