A subset of math problems I anticipate students will struggle with on the New GRE involve multiple answer questions. As you can see below, there are 8 possible answer choices, any of which can be correct. In many ways, this can be more devious than a numeric entry question.

*A group of x applicants is to be chosen for a spot on a television show. The number of applicants chosen, y, can range from 1 to x. If there are a total of 56 different possible selections for the spot on the television show, which of the following are possible values of y? (Choose all that apply.)*

*(A) **1*

*(B) **3*

*(C) **5*

*(D) **7*

*(E) **8*

*(F) **28*

*(G) **55*

*(H) **56*

I’m guessing most will miss this problem, as it is very tricky. The good news is we can walk through it together, and apply the approach to future questions.

The first thing many students do—and something you should avoid doing—is try to create an equation. Even if you know the combinations/permutations formula, trying to set up a formula with x and y can be far more challenging and time consuming than plugging in and working with the answer choices.

We know there is a total of 56 different teams. So, let’s play with the answer choices. I’ll start near the middle with 5.

Let’s say I choose 3 from the 5. Will that give me 56?

I use a combinations formula because I’m choosing a small group from a large group. Using the numbers above, I get 5!/3!2! = 10. That is far too low. I don’t want to choose 4 or 1 for y, because that will give me a number that is even lower. With combinations the closer together the two numbers in the denominator (in this case 2 and 3), the greater the result of the combinations equation.

Next I plug in 8. It’s generally a more friendly number than 7, which I tend to avoid. I’ll assume that y is equal to 5 (again, at this point, I don’t want to plug in a number that is too close to x or too low, because that will give me a low number).

8!/5!3! gives me 56.

Now, here is a little twist that applies to all combinations. There are usually two answers that yield the same solution. Let’s say I’d chosen 3, instead of 5. What would my equation have looked like? 8!/3!5!. This is the exact same as 8!/5!3!, both of which give us: 56.

Now I’ve found two y values that work for the question. Are there any others? The good news is I don’t have to plug in a bunch of numbers, as there are only a few possibilities left. We can get rid of 7!/4!3! = 35, because it is too low.

Answer Choice (F) 28 is a good place to start. Let’s make sure that y, the number we are choosing from, is very low or very high. I will choose either 2 or 26, both of which give me the equation, 28!/2!26! If I had chosen a number somewhere in the middle, such as 10, I would end up getting a very large number. Even if I choose a very low number, or a number close to 56, I end up getting a number much greater than 56, such as 28!/2!26! = 28 x 27 divided by 2. However, if we chose a lower number or a number closer to 28, such as 27, then I would get 28!/27!1! = 28.

Ah, so if I plug in a number y, that is one less than x, or equal to 1, then that value is going to yield x. For instance, if I go back and choose the number 5 for x, and choose 1 less than 5, 4 for y, I end up getting 5!/4!1!, which equals 5. I could also choose 1, which will give me the same result.

So finally—and thanks for bearing with me, as these are some pretty advanced combinations—if want to yield 56 total different selections, then we have to have 56 members and choose either 1, (56!/1!55!) or 55 (56!/55!1!).

Now comes the tricky part—as if your head isn’t already spinning. We are not looking for x, which would be 8 or 56. We are looking for y. Going back to the beginning of the question, we had chosen both 3 and 5 for y when selecting from a group of 8. Therefore (B) 3 and (C) 5 are answers.

Next we had x equals 56. The only y values that resulted in 56 different selections were (A) 1 and (G) 55.

The final answer: (A), (B), (C), and (G).

I haven’t yet seen enough New GRE math questions to accurately rate the level of difficulty in this question…but if you got it right, give yourself a pat on the back. In my humble opinion, this is definitely a 700+ question.

Nice math 🙂 ….

So why would someone start with the numbers in the list as x values when they are meant to be possible y values? Are you starting like this due to you knowing the answer and how to approach it? And 1 to x would be a better way to phrase it?

We plug in possible Y values as X values because X represents the upper boundary of the numerical range for Y values. You want to check for the highest value of Y among the answer choices that yields 56 when it’s used as an X value; in other words, you want to find the Y value among the answers that comes closest to being X. If you can find the answer choice for Y that comes closest to being X, you’ll have the top part of your range– this can help you eliminate answers that are too big, and can help you start to plug in other Y values that are smaller than X.

1 to X would be another way to phrase it— it can be phrased either way, as we’ve already established that X is the maximum number in the range for Y. X to 1 suggests the range could be as much as larger number X, but then moves on down to 1 as its minimum. But 1 to X– smallest number to largest– is more conventional wording. So wording it that way may be more helpful. Thank you for bringing this to our attention— I’ve just edited the wording so that it now reads “1 to X.”

This can be solved easily if we use pascal’s Triangle

Great brain teaser!

However, I do not understand the way you exclude the answer choices D and E.

I personally solved it like this:

xC3 = 56

x(x-1)(x-2)/6 = 56

x(x-1)(x-2) = 336

three consecutive integers with product 336 yields 8*7*6 so we have 8C3 = 56 and necessarily 8C5 = 56. 56C1 and 56C55 were quite obvious, and the fact that xC26 would be too large as well.

However, I cannot think of a way to conveniently exclude D and E and I do not understand why do you check for 7C4 at this point: “Are there any others? The good news is I don’t have to plug in a bunch of numbers, as there are only a few possibilities left. We can get rid of 7!/4!3! = 35, because it is too low.”

The only way I can think of is by checking if 9C7 and 10C8 (the most plausible candidates) are equal to 56 but this does not seem rigorous.

I was just brushing up on my combinations/permutations when I came upon this question. I may be wrong, but shouldn’t the question be:

‘A group of x applicants is to be chosen FROM for a spot on a television show…’? since we’re working out for xCy?

Yeah, it is worded a little strangely. I think a clearer way would have been as follows:

x applicants hope to appear on a T.V show. The number of applicants chosen, y, …

Sorry for any confusion!

Absolute mind bender… Thanks for sharing this question Chris….

Yeah, this question is definitely a toughie!

I would suggest that the way this question is asked is rather vague, which adds to its difficulty, IMHO.

“a total of 56 different possible selections for the spot” I believe would be better written as “56 different ways that the candidates could be selected for the spots available”

“what could be the value of y” would be better written “which of the following values are possible values of y.”