a) How many choices are possible if both books are from the same subject:
initially I approached the problem as follows:
${6 \choose 2} \times {7 \choose 2} \times {4 \choose 2} \to \frac{6!}{2!4!} \times \frac{7!}{2!5!} \times \frac{4!}{2!2!}= 15 \times 21 \times 6= 1890$
However the back of the book says there are exactly 42 ways, upon looking at my work and trying to see as to how it is 42 it appeared to me that instead of multiplying $15,21,6$ I should have added them together which comes up to 42. The question that I have is why in this instance was I supposed to add my results? Just about every other question involves multiplying.
b) What if the books are to be on different subjects?
we have 7 choose one, 6 choose 1, 4 choose 1, out of these we have 3 choose 2. I am not sure how to go about approaching this and the book says there are 94 choices.
Best Answer
Although I'm not that familiar with combinatorics, I believe what you're calculating, when finding the product of $\binom{6}{2}\binom{7}{2}\binom{4}{2}$ , you're calculating the amount of possibilities to choose a total of 6 books, of which 2 books should be Maths books, 2 Science and 2 Economics.
Instead, do this: your problem naturally divides into three problems: you're selling 2 Maths books, 2 Science books, or 2 Economics books. Notice the or, which in combinatorics often means that you should add these subamounts to get the total amount. Thus: you can sell two Maths books in $\binom{6}{2}$ ways, and similarly, $\binom{7}{2}$ for Science and $\binom{4}{2}$ for Economics. You have to make a conscious choice which category of books to sell, so you can't sell books from different categories, thus the total amount of possibilities is the sum of the possibilities per category.
As for your problem b), once again we can divide this into different problems: let us define
ato be a random Maths book,ba random Science book,ca random Economics book. Then we can do the following: we have{a,b}, or{a,c}, or{b,c}(for example{a,b}symbolises choosing one Maths book and one Science book).This gives us:
{a,b}has $6\cdot 7$ possibilities{a,c}has $6\cdot 4$ possibilities{b,c}has $7\cdot 4$ possibilities.We may once again add these, as they're the subtotals of the total problem, which gives us $94$.