PS : Probability # 6... History...and a Doubt

i can see what u are saying.
praet i always solve probability and combinations use basic fundamentals.
Just to revise:
probability of event X = Number of favourable events that results in X/ Total number of possible outcomes.

Now, two find above two quantities whether u use combinations or permutations really does't matter as far as you know what u are calculating. (i will take your example)

1) we have 5 History books , 7 Math books..
What is the probability that two books selected are on the same subject?

Total outcomes = 12C2 ------------ a
Favourable outcomes = 5C2 + 7C2 --------------- b

Thus, probabiliry = eq(b)/eq(a)
= 5*4/12*11 + 7*6/12*11 (as you got)

2) prob of getting 3 heads in 6 tosses of a die? (errata: read die as coin)
Total outcomes = 2^6 ---------- a
Favourable outcomes = 6C3*1 = 6P3*3! -------- b

Thus, probabiliry = eq(b)/eq(a)
= 6C3/2^6

The point that i am trying to make is that stick to the basic formaula for calculating and follow the counting methods or permutations (which ever applicable to the question asked) to get a and b (as stated above)
and u will be able to solve any probability problem with ease. Its very important not force the solutions of probability questions since a slight variation in actual test may lead to problem/mistakes.
As for your original question again use the above basics:
Part 1
Q: We have 4 History, 2 physics and 6 chemisty books. If we draw 6 books, what is the probability that we will have atleast 2 history and 1 chemistry book amongst them?
Total ways = 12C6 ----------a
Number of favourable ways = 4C2*6C1*9C3 ----------b

Thus probability for above = eq(b)/eq(a)

(Note: i have assumed that all books are different, if books of one type is identical then situation would be different)
hope this clarifies...
- Vicks