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Source: Official Guide Revised GRE 1st Ed. Part 8; Section 5; #11


A reading list for a humanities course

A reading list for a humanities course consists of 10 books, of which 4 are biographies and the rest are novels. Each student is required to read a selection of 4 books from the list, including 2 or more biographies. How many selections of 4 books satisfy the requirements? 90, 115, 130, 144, 195

3 Explanations


Maggie Glasser

Can you please explain how you get 115 from 4 bios and 4 bios?

Mar 5, 2019 • Comment

Sam Kinsman

Hi Maggie,

I'm happy to help - but I'm not quite sure I understand what you are asking about.

115 is the final answer - are you asking how we got that? If so, you might want to rewatch the explanation video that's below, and let us know if you have a question about a specific step in the video.

Also, I'm not quite sure what you meant by "4 bios and 4 bios." Perhaps that was a typo?

If you can tell us a bit more about where you're getting stuck, we can help you further!


Mar 6, 2019 • Reply


Why can't you do 4C2*8C2? (number of ways to choose 2 from the list of biographies multiplied by the number of ways to choose 2 from all the books remaining, including the 2 biographies that weren't selected)

Aug 24, 2014 • Comment

Lucas Fink

You have a clever idea in approaching this question, but there's one flaw: you separated the two books chosen at the beginning (the two biographies) from the books chosen later, but there may actually be some overlap between those groups. That is, the first two biographies you pick to fulfill the "at least two" are actually interchangeable with any biographies you pick later because this is all one group, but you've separated them from the other biographies. When we count the options for combinations of biographies, we need to keep them together, not put them into two different combination calculations.

To illustrate what I mean by that, let's look at a few different possible book choices. Let's say you pick these books:

B1 B2 | N1 N2 N3 N4

The left side is a possible combination from your 4C2 and the right side is an outcome of your 8C4. That's the same as picking these books:

B2 B1 | N1 N2 N3 N4

Or these:

B1 B2 | N4 N3 N2 N1

The order doesn't matter within each group, and that's just fine--you accounted for that by using the combination formula. But what if we pick a biography on the right side?

B1 B2 | B3 N1 N2 N3
B2 B1 | B3 N1 N2 N3
B1 B2 | N3 N2 N1 B3

Those are all the same, right? And you accounted for them all being the same. The repetitions on the left side are accounted for by your first calculation, and the repetitions on the right are accounted for by your second calculation. You're only counting this as one result, as you should. But what about this one?

B1 B3 | B2 N1 N2 N3

Now we have a problem. The left side (B1 B3) was a completely separate result from picking B1 and B2, so that was considered a distinct combination by your first calculation. Similarly, the right side (B2 N1 N2 N3) is different from our previous right side (B3 N1 N2 N3), so your calculation left that option as distinct. But the total combination s actually no different.

B1 B2 B3 N1 N2 N3
B1 B3 B2 N1 N2 N3

In short, we need to count the combinations of biographies separately from the combinations of novels because we are making a distinction between them.

Aug 26, 2014 • Reply

Kelvin DeCosta

Hey if you still didn't get it:

Take this case:
B1 B2 and 2 other books.
If the 2 other books has a biography then there can be overlaps.

B1 B2 | B3 N1 gives the same list as B1 B3 | B2 N1.
This gets worse when all 4 are biographies. It is supposed to be 1 but you get:

B1 B2 | B3 B4,
B1 B3 | B2 B4,
etc. which is 12.

I tried the same thing approach and this is how I realized it was wrong.

Aug 17, 2019 • Reply


Chris Lele

Oct 11, 2012 • Comment

Arpit Mehta

Why can't we use the formula explained in tutorials? i.e counting formula starting with restrictive stages.

There are 4 stages and starting with most restrictive stage, there the 1st stage will have 4 (4 ways of choosing B), 2nd stage will 3 (one B out of remaining 3), 8 ways of choosing 3rd stage (2B and 6N) and 7 ways of choosing last stage (7 of remaining books) .The combination then is 4*3*8*7.
Obviously its wrong answer, but I don't what am I missing here.

Nov 17, 2013 • Reply

Lucas Fink

Hi Arpit,

Good question—Chris wasn't joking when he said that this is really tough problem.

The first thing to note is that we do not actually have separate stages in choosing the books. Let's call the biographies b1, b2, b3, and b4. Now, if I pick b1 and b2, that's the same as choosing b2 and b1.

For that reason, we need to use combinations, not the FCP, as you tried. If we use the FCP, then we are not considering the redundancy. This can be a very subtle distinction.

The second issue is a bit less subtle. The questions asks for the ways we can get AT LEAST two biographies. That means we also need to consider the combinations possible with three or four biographies.

If you have any further questions on this, you can reach out via the black "help" tab on the left side of the page :-)


Nov 22, 2013 • Reply

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