# Confusion over Combinations and Permutations

Just when I thought I understood everything, I have yet again made myself confused and cannot resolve this issue. Consider selecting 3 people from 5 where the order of selection matters, this is clearly permutations and is equivalent to $$5\times4\times3$$. However, if the order doesn't matter we must divide by $$3!$$ as each of these cases were initially treated differently but are all the same, when considering combinations. Makes sense.

However, now consider the number of ways of selecting a book and newspaper from 6 books and 3 newspapers. Clearly this problem is again combinations, and can be calculated as $$6\times3$$, yet this time we don't have to divide for some reason. We are dealing with combinations like the previous example again, but this time we don't divide by the number of ways to arrange the book and newspapers $$2!$$. Why is this the case?

• In first case you have $1$ type of objects, people, from where you make choice. In second there are $2$ type objects - books and newspapers. Commented Apr 21 at 7:15
• Suppose you had to select 3 books and 2 newspapers from the 6 books and 3 newspapers? Commented Apr 21 at 7:34

You are (in effect) dividing to accommodate that you are computing combinations, rather than permutations. Here, it is assumed that regardless of whether you are computing combinations or permutations, the order of selection: [book(s) then newspaper(s)] or [newspaper(s) then book(s)] is deemed irrelevant.

For illustrative purposes, first assume that you are selecting 2 books and 2 newspapers. Then, the number of possible permutations is

$$(6 \times 5) \times (3 \times 2),$$

and the number of possible combinations is

$$\frac{6 \times 5}{2!} \times \frac{3 \times 2}{2!}.$$

Now, revert to the original problem, where you are selecting exactly 1 book and 1 newspaper.

Then, the number of possible permutations is

$$(6) \times (3),$$

and the number of possible combinations is

$$\frac{6}{1!} \times \frac{3}{1!}.$$

In the second case, we have to select one book and one newspaper from 6 books and 3 books. Let the selection of books and newspapers be a separate event. If there are 'm' possible ways in event 1 and 'n' possible ways in event 2 then the number of possible ways of selection is (m x n) ways. Now number of possible ways of selecting 1 book from 6 books is 6 and the number of possible ways of selecting 1 newspaper from 3 books is 3. So the total ways possible is 6x3. If the question asked is to find the number of ways of selecting two books from 6 books and one newspaper from 3 newspapers you have to find the number of possibilities of selection of books which is 6!/2!(6-2)! and multiply this with the number of possibilities of selection of newspapers which is 3!/1!(3-1)!=3.

• The first thing to remember is that if more than one category is involved, we need to compute for each category separatelybefore combining.

• So in your example of books and newspapers, you have to compute separately for each before combining, and you have divided, unknowingly, by $$1!$$ for each category, i.e. you have computed $$\frac{6}{1!}\cdot\frac{3}{1!}$$

• Had you needed to choose $$2$$ books and $$3$$ newspapers, the computation would clearly show the division,viz $$\large\frac{6\times5}{2!}\times\large\frac{3\times 2\times1 }{3!}$$

• Finally, why do you need to compute permutations and divide for such problems when there exists a "choose function" for combinations that gives you the answer directly ? In the last example, and using the choose function, possibly written in your textbook either as $$^6C_2\times^3C_3,\quad C(6,2)\cdot C(3,3)\quad or \quad \binom63\binom33$$