# combinatorial analysis basics

(From Sheldon Ross, First course in probability - p. 17, problem 16)

A student has to sell 2 books from a collection of 6 math, 7 science and 4 economics books.

How many choices are possible if
(a) both books are to be on the same subject
(b) the books are to be on different subject?

Answer for part (b): There are 6⋅7 choices of a math and a science book, 6⋅4 choices of a math and an economics book, and 7⋅4 choices of a science and an economics book. Hence, there are 94 possible choices.

My logic is - 1 math book out of 6 can be selected in 6 ways and second book out of (7 science + 4 economics) in 11 ways, so total 66 OR 1 science book out of 7 in 7 ways and second book out of (6 math + 4 economics) in 10 ways, so total 70 ways OR 1 economics book out of 4 in 4 ways and second book out of ( 6 math + 7 science) in 13 ways, so total 52 ways, adding the 3 numbers we get total = 66 + 70 + 52 = 188.

Where am I going wrong?

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+1 for showing your thinking. It makes the answers more specific and easier to give. –  Ross Millikan May 24 '11 at 17:33

It can be very good strategy to deliberately double-count, then divide the answer you get by $2$.

But in this case the approach suggested by the book's answer is probably best. The order in which the books you are selling are chosen does not matter. So imagine them arranged on a shelf in order of importance, the math books first, then the science books, then the economics books.

If you have to sacrifice a math book, it can be chosen in $6$ ways. For each such choice, there are $7$ ways to choose the accompanying science book, for a total of $6\cdot 7$, and $4$ ways to choose the economics book, for a total of $6\cdot 4$. If you do not choose to sacrifice a math book, the only possibility is to pick science then economics, for a total of $7 \cdot 4$. Now add up.

Or else one can use your shortcut. If you choose to sacrifice a math book, then for each choice of a math book you can choose the "other" book in $11$ ways, for a total of $6 \cdot 11$. Then there are the $7 \cdot 4$ science and economics possibilities.

In this analysis, imagining lining up the books is important. It makes the process concrete.

Another way: Let's assume you have successfully solved part (a), and found the number $s$ of ways to choose $2$ books on the same subject. I get $s=42$. There are $C(17,2)$, that is, $136$, ways to choose $2$ books, with no restrictions. So there are $136 -42$ ways to choose the $2$ books if they are to be in different subjects.

However you decide to find the answers to (a) and (b), the sum of the answers should be $C(17,2)$.

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I think the second solution is more elegant. I also used the same approach and got the answer (94). –  Obinna Okechukwu May 24 '11 at 17:29

You're counting everything twice. For example, you're counting math and science as math and science-or-economics, but you're also counting it as science and math-or-economics.

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