Faulty Combinatorial Reasoning? I have 10 books, 4 of which are biographies while the remaining 6 are novels.  Suppose I have to choose 4 total books with AT LEAST 2 of the 4 books being biographies.  How many different combinations of choosing 4 books in such a way are there?
The following line of reasoning is faulty, but I can't figure out why:
First we figure out how many ways there are of choosing 2 biographies from 4.  Then we multiply this by the number of ways there are of choosing 2 of any of the remaining books from 8.  This way we will ensure that we get at least two biographies (perhaps more) when we enumerate the choices.  Then we have:


*

*BIOGRAPHIES: There are (4*3)/2! choices for the two biographies (we divide by 2! since the order in which the two biographies are chosen doesn't matter).

*REMAINING BOOKS: There are now 8 books left (6 novels, 2 biographies), which can be chosen in any order.  This leaves us with (8*7)/2! choices.

*Overall we have [(4*3)/2!]*[(8*7)/2!] = 168  total choices.


Where did I go wrong?
 A: In your reasoning, you are counting some cases several times. For example, if you take the biographies $B_1$ and $B_2$ as your mandatory biographies and take $B_3$ and $B_4$ as the two other ones, or if you take $B_£$ and $B_4$ as the mandatory ones and $B_1$ and $B_2$ as the other books, it is the same choice of $4$ books, but it will be counted twice.
To solve the problem:


*

*the number of ways of choosing $4$ books is $A_4 = \frac{10!}{4!\times6!} = 210$

*the number of ways of choosing $4$ books with no biographies is $B_0 = \frac{6!}{4!\times2!} = 15$

*the number of ways of choosing $4$ books with exactly $1$ biographies is $B_1 = 4\times\frac{6!}{3!\times3!} = 80$ (you pick $1$ biography amongst $4$ and then choose $3$ novels).

*the number of ways of choosing at least two biographies is $B_2^+ = A_4 - (B_1+B_2) = 115$.   

A: Your enumeration would count the following possibilities (and many other similar examples) separately:


*

*First choose Bio1 and Bio2, then choose Bio3 and Novel1.

*First choose Bio1 and Bio3, then choose Bio2 and Novel1.


Notice the only difference is that I switched when Bio2 and Bio3 would be chosen.
In general, it is best to split "at least"-type questions into several instances of "exactly". Here, you should try to count the ways to get exactly two, exactly three, and exactly 4 biographies as separate problems, then add up all the cases.
A: Suppose the biographies are of $A$, $B$, $C$, and $D$. Among the ways you counted when initially you chose two biographies, there were the biographies of $A$ and of $B$.  Among the choices you counted when you chose two more books was the biography of $C$ and novel $N$.  So among the choices counted in your product was choosing $A$ and $B$, then choosing $C$ and $N$.  That made a contribution of $1$ to your $168$. 
But among the choices you counted when you chose two biographies, there were the biographies of $A$ and $C$. And among your "two more" choices, there was the biography of $B$ and novel $N$. So among the choices counted in your product, there was the choice of $A$ and $C$, and then of $B$ and $N$. That made another contribution of $1$ to your $168$. 
Both of these ways of choosing end us up with $A$, $B$, $C$, and $N$.  So does choosing $B$ and $C$ on the initial choice, and $A$ and $N$ on the next.  Still another contribution of $1$ to your $168$.
So your product counts the set $\{A, B, C, N\}$ three times.  This it does for every combination of three biographies and one novel. It also overcounts the set $\{A, B, C, D\}$.
One could adjust for the overcount. In some problems that is a useful strategy.  Here it takes some care.
But a simple way to solve the problem is to count separately the ways to choose two bios, two novels; three bios, one novel; four bios, no novels and add up.
A: When you do this you sum two times the cases when you have more than two biographies books. Then need to do this
2 BIOGRAPHIES books: $\dfrac{4\cdot3}{2!} \cdot \dfrac{6\cdot5}{2!}=90 $
3 BIOGRAPHIES books:   $\dfrac{4\cdot 3 \cdot 2}{32!} \cdot 6 =24$
4 BIOGRAPHIES books: 1
then the result is 115.
