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.