# The Hardest GRE Math Question I've Ever Seen

#### 5/25/2015

#### YOU'RE GOING TO GET THIS QUESTION WRONG.

**Ok - I admit it - I was just trying to get your attention with that last line.** But seriously - I've never had a student who has answered the following actual GRE question correctly. It's the hardest GRE math question in the Official Guide, and I couldn't figure it out the first time I saw it, either. Let's check it out and then see why you should care about math challenge questions:**A certain bookcase contains ten books. Four are biographies and six are novels. A student must select four books; however, two or more of her selections must be biographies. How many different ways are there for the student to select four books?**

(The solution is at the end of the newsletter.)

The reason I think this one is so hard is that, although it is testing concepts with which students may be familiar, those concepts are presented differently than they'd be in math classes or textbooks. Welcome to the GRE - knowledge of concepts isn't sufficient. *Mastery***of those concepts is what's required.**

Before I explain the question (please try it if you're so inclined), let me explain why hard questions matter. Here are three ideas:**1. They force a student to think. In doing so, the student builds math ability.**

2. If a student can solve them, his math confidence goes up.

3. They get students used to adapting - a crucial skill for the GRE.

It's *so important* that students develop a "can do" attitude towards math. They don't even need "challenge" questions. What I'd recommend is this: **do notÂ look up the answer or explanation to a question until you've truly tried your best on it**. This will build your skill and confidence. If you can't get it, fine, but remember - as soon as you stop thinking and look up the answer, you're missing a chance to strengthen your math muscles and confidence.

**Ok - here's the solution:**

There are three scenarios to consider here. In the first, the student picks all four biographies. There is only **one way** to do this.

In the second, the student picks three biographies. To find out how many ways the student can select 3 from a group of 4, you can use the **groups formula**:****(4 * 3 * 2 * 1)-------------------(3 * 2 * 1) (1)**

This reduces to four ways. Or, you could write out the possibilities, using A, B, C, and D to represent each biography.

The next step in this second scenario is to figure out how many ways there are to choose one novel from six (since you're choosing three biographies, you must also pick a novel). Logically, there are six ways to pick one thing from a group of six.

Finally (for this scenario), multiply the four ways to choose three biographies by the six ways to choose a novel to get

**24 additional ways**.

Now... the third scenario: picking two biographies and two novels. To figure out how many ways there are to pick two biographies from four, we'll use the groups formula:

**(4 * 3 * 2 * 1)**

------------------

(2 * 1) (2 * 1)

------------------

(2 * 1) (2 * 1)

Which reduces to 6 ways to pick 2 from 4. Now, we'll do the same with the novels - to find how many ways to pick two from six, it's the groups formula again:

**(6 * 5 * 4 * 3 * 2 * 1)**

-----------------------------

(2 * 1) (4 * 3 * 2 * 1)

-----------------------------

(2 * 1) (4 * 3 * 2 * 1)

Which reduces to 15: 15 ways to pick 2 from 6. Finally, multiply 15 ways to pick the two novels by 6 ways to pick the two bios and get

**90 additional ways.**

To get to the final answer, add the 1 initial way plus the 24 ways from the 2nd scenario plus the 90 ways from the 3rd scenario,

**and your answer is 115 ways.**

Hope you enjoyed that! See you in a couple of weeks. What about you - do you agree that it was the hardest GRE math question you've seen?

* If you got this question right on your own, please tell me and I will be impressed.

** the "groups" formula is used when you want to find out how many groups of a certain number you can make from a larger group. For example, if you wanted to figure out how many groups of 3 you could make out of a group of 8. The arrangement of the group doesn't matter in this kind of problem (or we'd have to use a different method): the group ABC is the same as CBA or BAC. I'd define the formula as:

**(big number)!**

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(little number)! (big number - little number)!

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(little number)! (big number - little number)!