If two of the four expressions x + y, x + 5y, x - y, and 5x - y are

Okay, first recognize that x² - (by)² is a DIFFERENCE OF SQUARES.
Here are some examples of differences of squares:
x² - 25y²
4x² - 9y²
49m² - 100k²

In general, we can factor differences of squares as follows:
a² - b² = (a-b)(a+b)

So . . .
x² - 25y² = (x+5y)(x-5y)
4x² - 9y² = (2x+3y)(2x-3y)
49m² - 100k² = (7m+10k)(7m-10k)

----OKAY, ONTO THE QUESTION------------------

So, from the 4 expressions (x+y, x+5y ,x-y and 5x-y), only one pair (x+y and x-y) will result in a difference of squares when multiplied.

So, the question now becomes:
If 2 expressions are randomly selected from the 4 expressions, what is the probability that x+y and x-y are both selected?

P(both selected) = [# of outcomes in which x+y and x-y are both selected]/[total # of outcomes]

As always, we'll begin with the denominator.

total # of outcomes
There are 4 expressions, and we must select 2 of them.
Since the order of the selected expressions does not matter, we can use combinations to answer this.
We can select 2 expressions from 4 expressions in 4C2 ways (= 6 ways)

If anyone is interested, I have added a video (below) on calculating combinations (like 4C2) in your head

# of outcomes in which x+y and x-y are both selected
There is only 1 way to select both x+y and x-y

So, P(both selected) = 1/6

Answer: E

RELATED VIDEO

The only way to get x^2 - (by)^2 is if you multiply (x+by) with (x-by). Looking at the expressions given, the only pair that will satisfy this is (x+y) & (x-y)

No of ways to pick pairs = C(4,2) = 6

No of pairs satisfying condition = 1

Probability = 1/6

How do I know from the phrasing of this problem that, once chosen, an expression is taken off the list (ie not replaced)? When I was solving the problem I assumed that an expression of form (x+y)(x+y) is possible. In that case the number of combinations is 8.

Question says: "If two of the four expressions x+y, x+5y, x-y, and 5x-y are chosen at random..." so, at least for me, it's quite obvious that there is no replacement whatsoever. Also if it were the case the question would explicitly mention that.

How come we are assuming that we will choose both expressions at once? That is, why not assume that we choose one of the four expressions, then choose the other one?

For example, 4 letters numbered 1 2 3 4, two of which are chosen at random, what is the probability that the chosen letters will be odd numbered?

Don't these two problems have the same concept?

Mathematically the probability of picking two expressions simultaneously (at once), or picking them one at a time (without replacement) is the same.

Hello Bunuel,

Maybe telling you how I solved the problem (and got a wrong answer) will explain how I was confused.

Since the wanted product is (x + y ) (x - y), I assumed that I will be picking (x+y) then (x-y) [from the expressions) which means choosing 1/4 then choosing 1/3 and multiplying them to get 1/12 which is wrong. I think I approached it the same was as 144144 (posted above) did.

Thank you very much Bunuel.

It should be 1/4*1/3*2, since you can choose (x+y) first and then (x-y) or vise versa.

VERY SIMPLE WAY TO SOLVE THIS PROB WITHOUT ANY COMPLEX THEORY N ANY CALCULATION :
FROM THE QUESTION STEM , YOU COULD UNDERSTAND VERY EASILY THAT ONLY ONE COMBINATION I.E (X+Y) AND (X-Y) IS THERE TO GET X2-(BY)2
SO WE HAVE TO CHOOSE ONLY (X+Y) AND (X-Y)
wE CAN SELECT IT IN TWO WAY I.E . (X+Y) AND THEN (x-Y) OR (X-Y) AND THEN (X+Y)
FOR FIRST CASE PROBABILITY =(1/4*1/3)
FOR SECOND CASE PROBABILITY =(1/4*1/3)
sO TOTAL PROBABILITY =2*((1/4*1/3)=1/6
HOPE IT COULD SIMPLY WORK FOR YOU

the result will be of form : \(x^2 - (by)^2\) only when we choose x+y and x-y...rest all pair will not satisfy.
therefore favorable case=1
total case = we have to select 2 from 4 ....= \(4C2\) = 6

probability = favorable/total =\(1/6\)
hence E

Hi Bunuel,

Here is how I did it. Since we can choose either (x+y) or (x-y) at first, the probability is 2/4 and for second pick we have only one of these left so the prob is 1/3. Total prob= 2/4*1/3=1/6
Is that right?
Thanks!

First, notice that we are being tested on the difference of squares. We can restate the problem as: What is the probability, when selecting two expressions at random, that the product of those expressions will create a difference of two squares? Remember, the difference of two squares can be written as follows:

a^2 - b^2 = (a + b)(a - b)

So, x^2 - (by)^2 can be written as (x + by)(x - by). Thus, we are looking for two expressions in the form of (x + by)(x - by). Although this problem is attempting to trick us with the expressions provided, the only two expressions that when multiplied together will give us a difference of squares are x + y and x - y. When we multiply x + y and x - y, the result is x^2 - y^2 or x^2 - (1y)^2.

We see that there is just one favorable product, namely (x + y)(x - y). In order to determine the probability of this event, we must determine the total number of possible products. Since we have a total of four expressions and we are selecting two of them to form a product, we have 4C2, which is calculated as follows:

4C2 = (4 x 3)/(2!) = 12/2 = 6 products

Of these 6 products, we have already determined that only one will be of the form x^2 - (by)^2. Therefore, the probability is 1/6.

The answer is E.

Note: If you don’t know how to use the combination formula, here is a method that will work equally well:

We are choosing 2 expressions from a pool of 4 possible expressions. That is, there are 2 decisions being made:

Decision 1: Choosing the first expression

Decision 2: Choosing the second expression

Four different expressions are available to be the first decision.

For the second decision, 3 remaining expressions are available because 1 expression was already chosen. We multiply these two numbers: 4 x 3 = 12.

The final step is to divide by the factorial of the number of decisions (2! = 2) because the order in which we multiply the expressions doesn’t matter (for example, (x+y)(x-y) = (x-y)(x+y)). In this case, the two expressions are only considered as one, so we need to divide 12 by 2.

12/2 = 6

Once again the answer would be E.

Alternate Solution:

One other way to solve this problem is to use probability.

Once again, we have determined that the only two expressions that when multiplied together will give us a difference of squares are x + y and x - y. If we select either of those expressions first, since there are 2 favorable expressions and 4 total expressions, there is a 2/4 = 1/2 chance that either x + y or x - y will be selected. Next, since there is 1 favorable expression left and 3 total expressions, there is a 1/3 chance that the final favorable expression will be selected.

Thus, the probability of selecting x - y and x + y is 1/2 x 1/3 = 1/6.

Answer: E

Hi All,

This question is based heavily on algebra patterns. If you can spot the patterns involved, then you can save some time; even if you can't spot it though, a bit of 'brute force' math will still get you the solution.

We're given the terms (X+Y), (X+5Y), (X-Y) and (5X-Y). We're asked for the probability that multiplying any randomly chose pair will give a result that is written in the format: X^2 - (BY)^2.

Since there are only 4 terms, and we're MULTIPLYING, there are only 6 possible outcomes. From the prompt, you should notice that the 'first part' of the result MUST be X^2....and that there should be NO 'middle term'....which limits what the first 'term' can be in each of the parentheses....

By brute-forcing the 6 possibilities, you would have...
(X+Y)(X+5Y) = X^2 + 6XY + 5Y^2
(X+Y)(X-Y) = X^2 - Y^2
(X+Y)(5X-Y) = X^2 + 4XY - Y^2
(X+5Y)(X-Y) = X^2 + 4XY - 5Y^2
(X+5Y)(5X-Y)= 5X^2 +24XY - 5Y^2
(X-Y)(5X-Y) = 5X^2 -6XY + Y^2

Only the second option is in the proper format, so we have one option out of six total options.

Final Answer: [spoiler]E[/spoiler]

GMAT assassins aren't born, they're made,
Rich

hey, can you please help me in this question
i solved it by knowing out of 2 pairs we need 1 pair so 2c1 and out of 4 no. we need 2 no. so 4c2 so i got probability 2/6 which is 1/3 so how you get 1 in numerator. please explain thankyou

Hi rishabhmishra,

If you're going to treat this prompt as a Combination question, then that's fine (although it's not necessary). You just have to make sure that you treat the ENTIRE prompt correctly. Using the Combination Formula, you correctly determined that there were 6 possible "products" that could be created. However, you made a mistake in terms of what fraction of those 6 products actually 'fit' what you were looking for. The point of using the Combination Formula is that it is meant to eliminate 'duplicate entries', but it's not correct to use 2C1 here, since there was only ONE product that fit what the question asked for: (X+Y)(X-Y) and since that's the SAME product as (X-Y)(X+Y), we're NOT allowed to count it twice.

GMAT assassins aren't born, they're made,
Rich

1. The six possible products are:
(x+y)(x+5y)
(x+y)(x-y)
(x+y)(5x-y)
(x+5y)(x-y)
(x+5y)(5x-y)
(x-y)(5x-y)

2. Of the above, we see only the second gives the form required.
Probability is therefore 1/6

Why is the proper approach for this problem to set up a combination 4C2 where as for other problems such as choosing a team you may do 4 * 3. When I solved this problem I almost thought of the four expressions as players for a team. There are four to choose from and I have two slots to put players in. For the first slot, I have four to choose from and in the second slot, I have 3 to choose from so there are 4* 3 = 12 combinations. If there were replacement, that would yield 4 * 4 = 16 combinations.

I guess my question is: would someone be able to explain to me the flaw in my logic between these two different methods of arriving at the number of combinations. Thanks!