A certain panel is to be composed of exactly three women and

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

\(C^3_x=\frac{x!}{3!(x-3)!}=\frac{x(x-1)(x-2)}{6}=56 \implies x(x-1)(x-2)=6 \times 56 = 6 \times 7 \times 8 \implies x=6\)

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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