we have the four numbers x+y,x+5y, x-y, x-5y and need to pick 2 number such as the product of the selected numbers is x^2 - (by)^2 i.e.
now we can select the numbers in 4c2 ways (2 numbers out of 4) = 6 ways
now the product should be x^2 - (by)^2 which means they can be factored as
(x-by)*(x+by)
so the pair is
x+y & x-y .... x-5y & x+5y

hence the prob is
2/6 = 1/3

Correct me if wrong ...

regards
sason

Rahul@gurome wrote:

emiflo wrote:Hi guys. This is a tricky one for me.

If two of the four expressions x+y,x+5y, x-y, x-5y are to be selected at random, what is the probability that their product is equal to x^2 - (by)^2 where b is an integer.

a.1/2
b.1/3
c.1/4
d.1/5
e.1/6

x^2 - (by)^2 = (x + by)(x - by), which clearly implies that the expressions cannot be x + 5y and x - 5y. So, out of 4 expressions, we are left with 2 expressions.
Total no. of possible combinations = 4C2 = 6

Therefore, the required probability = 1/6

The correct answer is[spoiler] (E)[/spoiler].

I didnt get you logic ... can you please explain
1.expressions cannot be x + 5y and x - 5y???
2. shouldnt the answer be 2/6 =1/3 ???

Rahul@gurome wrote:

emiflo wrote:Hi guys. This is a tricky one for me.

If two of the four expressions x+y,x+5y, x-y, x-5y are to be selected at random, what is the probability that their product is equal to x^2 - (by)^2 where b is an integer.

a.1/2
b.1/3
c.1/4
d.1/5
e.1/6

x^2 - (by)^2 = (x + by)(x - by), which clearly implies that the expressions cannot be x + 5y and x - 5y. So, out of 4 expressions, we are left with 2 expressions.
Total no. of possible combinations = 4C2 = 6

Therefore, the required probability = 1/6

The correct answer is[spoiler] (E)[/spoiler].

thanks......but i didn't get our logic can you explain me our logic clearly.........