Here’s a question I’m struggling with:

There are 10 book consisting of 4 biographies and 6 novels. A person stacks four of the books together. In the stack of four books, at least 2 books must be biographies. How many possible permutations are there for stacking the four books?

I thought of two ways to do this problem:

(# permutation for biographies with r=2) * (# permutations for 8 remaining books with r=2) * (# of possible positions where the two biographies can occupy)

$$ ^4 P_2 * ^8 P_2 * ^4 C_2= \\\dfrac{4!}{2!} * \dfrac{8!}{6!} *\dfrac{4!}{2!*2!}= \\4*3*8*7*6= 4032$$

The rationale behind this method is as follow:

The other method I used is as follow: to satisfy the requirement, you first pick two biography books out at random, which has 4*3 permutations. Then, out of the 8 remaining books, you pick 2 random books, which has 8*7 permutations. In terms of order, there are 6 total combinations for ordering of biography B and other books X (BBXX, BXBX, BXXB, XBBX, XBXB, XXBB). Thus the solution should be 4*3*8*7*6

The other method I used is as follow:

(# of total permutations) - (# of permutations with no biography) - (# of permutations with exactly 1 biography)

$$ ^{10} P_4 - ^6 P_4 - ^6 P_3 * ^4 P_1 * ^4 C_1 \\ \dfrac{10!}{6!}-\dfrac{6!}{2!}-\dfrac{6!}{3!}*4*4= \\ 5040-360-1920=2760 $$

The rationale behind this is simpler: from total amount of permutations, I subtract away permutations where no biography books exist and permutations where only one biography book exist, leaving only permutations with 2 or more biography books.

The two methods both make logical sense to me, so I’m lost as to why they give different results. I’m struggling to see what went wrong that would cause the two to have differing solutions

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    $\begingroup$ Your mistake is an incredibly common one. You overcounted in your first approach every scenario where you had more than two biographies selected by incorrectly applying some significance to whether or not a particular biography was selected in the first step versus selected as one of the "remaining" books in the later step. $\endgroup$
    – JMoravitz
    Commented Mar 11, 2020 at 17:39
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    $\begingroup$ You could fix this as you did in your second approach, or you could fix it by approaching directly counting the arrangements with exactly two biographies, adding this to the arrangements with exactly three, and so on... $\endgroup$
    – JMoravitz
    Commented Mar 11, 2020 at 17:40
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    $\begingroup$ This is a very good post, and especially good for a first post. I appreciate that you gave all of the information of everything you tried and specifically where you are confused. It makes it much easier to craft an answer that addresses the specific question(s) you have. $\endgroup$ Commented Mar 11, 2020 at 17:49

1 Answer 1


The first method is miscounting. Consider your first method of counting, and suppose you have all four biographies in the stack: $A,B,C,D$.

You are choosing $A,B$ from the biographies, then choosing $C,D$ from the remaining eight books, then permuting them. This is the same as choosing $C,D$ from the biographies, then choosing $A,B$ from the remaining eight books, and them permuting them.

You can modify the first method by counting the number of ways to stack the four with exactly 2 biographies plus exactly three biographies, plus exactly 4 biographies.

This calculation is the number of ways to choose four books for the stack, then permute the books. So, I am adding together the number of ways to choose 2 biographies and 2 novels, plus 3 biographies and 1 novel, plus 4 biographies, and then, finally, permuting them.

$$\left((^4C_2)(^6C_2)+(^4C_3)(^6C_1) + (^4C_4)(^6C_0)\right)4! = 2760$$

This gives the same answer you already found by your second counting method.


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