Thread: Combination problem involving book selling.

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

(b) the books are to be on different subjects?

So for the first pass one can choose one book out of one of the three piles so, $\displaystyle \binom{6}{1}+\binom{7}{1}+\binom{4}{1}=17$.

For the second pass one can only choose a book from a pool that wasn't selected from the first pool. Leaving 3 options;
1. If you selected from $\displaystyle \binom{6}{1}$ then $\displaystyle \binom{7}{1}+\binom{4}{1}=11$
2. If you selected from $\displaystyle \binom{7}{1}$ then $\displaystyle \binom{6}{1}+\binom{4}{1}=10$
3. If you selected from $\displaystyle \binom{4}{1}$ then $\displaystyle \binom{6}{1}+\binom{7}{1}=13$

I'm not entirely sure what I've done wrong here. I think I should just be able to add everything from pass one and pass two to get the correct answer but the listed answer is 94 and $\displaystyle 51\neq94$

Could someone clue me into what I'm doing wrong here?

Originally Posted by bkbowser

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
(b) the books are to be on different subjects?

I'm not entirely sure what I've done wrong here. I think I should just be able to add everything from pass one and pass two to get the correct answer but the listed answer is 94 and $\displaystyle 51\neq94$

Do it in one pass. You just have three possibles.

$\displaystyle (6)(7)+(6)(4)+(7)(4)=~?$.

Originally Posted by Plato

Do it in one pass. You just have three possibles.

$\displaystyle (6)(7)+(6)(4)+(7)(4)=~?$.

Huh. I didn't see that at all.

So in the first case you pull a book from the math and the science piles.
In the second case you pull a book from the science and economics piles.
In the third case you pull a book from the math and economics piles.

In any of the three cases the number outcomes is dependent on the pool sizes. So, taking the first case as an example, the math pile has 6 books and the science pile has 7 books; meaning that there are 6 different possibilities for the first pass and 7 possibilities for the second pass. Or, 6*7, outcomes.

Finally since each of the three cases is exclusive you add them all together.

Does this look right?

Originally Posted by bkbowser

So in the first case you pull a book from the math and the science piles.
In the second case you pull a book from the science and economics piles.
In the third case you pull a book from the math and economics piles.

Well they are just combinations.

$\displaystyle \binom{6}{1}\binom{7}{1}+\binom{6}{1}\binom{4}{1}+ \binom{7}{1}\binom{4}{1}=(6)(7)+(6)(4)+(7)(4)=~94$.