I assumed that X will have a unique value and got it right..But is this always true?

@ Edit:I just found this post by Bunuel..Seems relevant here too

Suppose we are told that there are 10 ways to choose x people out of 5. What is x? C^x_5=10 --> \frac{5!}{x!(5-x)!}=10 --> x!(5-x)!=12 --> x=3 or x=2. So we cannot determine single numerical value of x. Note that in some cases we'll be able to find x, as there will be only one solution for it, but generally when we are told that there are n ways to choose x out of m there will be (in most cases) two solutions of x possible.

But if we are told that there are 10 ways to choose 2 out of x, then there will be only one value of x possible --> C^2_x=10 --> \frac{x!}{2!(x-2)!}=10 --> \frac{x(x-1)}{2!}=10 --> x(x-1)=20 --> x=5.

In our original question, statement (1) says that there are 126 ways to choose 5 out of x+2 --> there will be only one value possible for x+2, so we can find x. Sufficient.

Just to show how it can be done: C^5_{(x+2)}=126 --> (x-2)(x-1)x(x+1)(x+2)=5!*126=120*126=(8*5*3)*(9*7*2)=5*6*7*8*9 --> x=7. Basically we have that the product of five consecutive integers ((x-2)(x-1)x(x+1)(x+2)) equal to some number (5!*126) --> only one such sequence is possible, hence even though we have the equation of 5th degree it will have only one positive integer solution.

Statement (2) says that there are 56 ways to choose 3 out of x+1 --> there will be only one value possible for x+1, so we can find x. Sufficient.

C^3_{(x+1)}=56 --> (x-1)x(x+1)=3!*56=6*7*8 --> x=7. Again we have that the product of three consecutive integers ((x-1)x(x+1)) equal to some number (3!*56) --> only one such sequence is possible, hence even though we have the equation of 3rd degree it will have only one positive integer solution.

Your examples are not relevant for this problem.

What I'm saying is that there is only one number out of which we can make 56 different selections of 3.

For example, if I say that we can choose 2 out of x in 10 ways, there would be only one x satisfying this, namely 5: xC2=10 --> x=5.

Hope it's clear.
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Hi,

I was training on this specific question on which I understand how to get to the DS answer although I did not get the explanation on how to make the calculation if this was a PS problem.

Here is the question:


A certain panel is to be composed of exactly three women and exactly two men, chosen from x women and y men. How many different panels can be formed with these constraints?

(1) If two more women were available for selection, exactly 56 different groups of three women could be selected.

(2) x = y + 1


Here is the explanation from Manhattan Prep:

In order to know how many panels we can form when choosing three women and two men, we need to know how many women and men we have to choose from. In this case, we need to know the value of x (the number of women to choose from) and the value of y (the number of men to choose from). The number of panels will be equal to the number of groups of three that could be chosen from x women multiplied by the number of groups of two that could be chosen from y men.

(1) INSUFFICIENT: This statement tells us that choosing 3 from x + 2 would yield 56 groups.

One concept that you need to know for the exam is that when dealing with combinations and permutations, each result corresponds to a unique set of circumstances. For example, if you have z people and know that choosing two of them would result in 15 different possible groups of two, it must be true that z = 6. No other value of z would yield exactly 15 different groups of two. So if you know how many subgroups of a certain size you can choose from an unknown original larger group, you can deduce the size of the larger group.

In the present case, we know that choosing three women from x + 2 women would yield 56 groups of 3. These numbers must correspond to a specific value of x. Do not worry if you do not know what value of x would yield these results (in this case, x must equal 6, because the only way to obtain 56 groups if choosing 3 is to choose from a group of 8. Since the statement tells us that 8 is 2 more than the value of x, x must be 6). The GMAT does not expect you to memorize all possible results. It is enough to understand the underlying concept: if you know the number of groups yielded (in this case 56), then you know that there is only one possible value of x.

(2) INSUFFICIENT: Knowing only that x = y + 1 tells us nothing specific about the values of x and y. Infinitely many values of x and y satisfy this equation, thus yielding infinitely many answers to the question.

(1) AND (2) SUFFICIENT: Taking the statements together, we know that (1) gives us the value of x and that (2) allows us to use that value of x to determine the value of y. Remember that, with data sufficiency, we do not actually need to calculate the values for x and y; it is enough to know that we can calculate them.

The correct answer is C.

Here is what I don't understand in the explanation and on which I would like clarity on:

- For example, if you have z people and know that choosing two of them would result in 15 different possible groups of two, it must be true that z = 6. No other value of z would yield exactly 15 different groups of two.

- in this case, x must equal 6, because the only way to obtain 56 groups if choosing 3 is to choose from a group of 8. Since the statement tells us that 8 is 2 more than the value of x, x must be 6

==> How do you get to those?

Thanks for your help.

Select 2 people from z people. We have \(C^2_z\) different ways.

\(C^2_z=\frac{z!}{2!(z-2)!}=\frac{z(z-1)}{2}=15 \implies z(z-1)=15 \implies z=6\)



Select 3 women from x woman, there are \(C^3_x=56\) different ways

\(C^3_x=\frac{x!}{3!(x-3)!}=\frac{x(x-1)(x-2)}{6}=56 \implies x(x-1)(x-2)=6 \times 56 = 6 \times 7 \times 8 \implies x=6\)
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Merging topics. Please refer to the discussion on previous pages. Your doubt is addressed there.

Also, please read carefully and follow: https://gmatclub.com/forum/rules-for-po ... 33935.html Thank you.
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