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I'm not understanding combinatorics usage between two questions, and I would like some clarity on it.


Question 1:

There are 10 books on a shelf, of which 4 are paperbacks and 6 are hardbacks. How many possible selections of 5 books from the shelf contain at least one paperback and at least one hardback?

In this question, I initially calculated each choice individually:

  • Pick one out of four paperbacks
  • Pick one out of six hardbacks
  • Finally pick three from the remaining eight

This led me to the formula: $4 \times 6 \times \binom{8}{3}$.

I then realised that this is duplicating selections. For example, call the paperbacks $[p_1, p_2, p_3, p_4]$ and the hardbacks $[h_1, h_2, h_3, h_4, h_5, h_6]$. In my formula, an allowed combination would be $p_1$ (one out of four paperbacks); $h_2$ (one of the 6 hardbacks); and $p_2, h_5, h_6$; this results in a selection outcome of $[p_1,h_2,p_2,h_5,h_6]$. However, another allowed combination would be p2 (one out of the four paperbacks); $h_5$ (one out of the six hardbacks); and $p_1, h_2, h_6$; this results in the same selection outcome of $[p_2,h_5,p_1,h_2,h_6]$ (as the ordering does not matter).

So clearly, this method of thinking about the problem is not correct. What I did was make each selection one by one (i.e. choose one out of four paperbacks, then one out of four hardbacks, and then three among the remaining eight), and this method seems to be wrong when each selection affects the selections coming after it. The answer for this question (if anyone is interested) is $(\binom{10}{5}) - (\binom{6}{5})$.

Now comes the next question..


Question 2

In how many different ways can a group of 8 people be divided into 4 teams of 2 people each?

The answer is provided as ($\binom{8}{2}\times \binom{6}{2}\times\binom{4}{2}\times\binom{2}{2})/4!$ and I kind of get why this is so, but I don't understand how this is any different. This method seems to me exactly as the one I did for the question above: make each selection one by one when each selection affects the subsequent selection space.


Now for my question: Why does the method work in the second case and not in the first case? In which situations can I select one by one? Kindly note: I want to understand the thinking or intuition behind using the method as opposed a mathematical, proof based argument that Stack seems to often provide.

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The methodological difference between the two questions is that the overcounting that you detected in your attempt for Question 1 is hard to control there but easy to control in Question 2.

In Question 2, choosing the teams one at a time (choosing 2 of 8 people for the first team, then 2 of the remaining 6 people for the second team, etc.) counts every possible division (into 4 teams of 2) exactly $4!$ times, because the 4 teams could have been chosen in any order. So the answer is obtained by taking the one-at-a-time count and dividing by $4!$.

In Question 1, on the other hand, the amount of overcounting depends on how many books of each sort were chosen. If you choose 1 paperback and 4 hardcovers, that set of books could have been chosen in 4 ways (because any one of the 4 hardcovers could have been chosen as the first one, before picking the remaining 3 as an unordered set. But with 2 paperbacks and 3 hardcovers, either of the 2 paperbacks could have been chosen first and any one of the 3 hardcovers could have been next. So you'd have counted this set of books 6 times. As a result, you can't go from the one-at-a-time count to the final answer by just dividing by the overcounting number; there isn't a single overcounting number.

Although it's not what you asked, let me also indicate how to solve Question 1. If you choose 5 books, you'll automatically have at least 1 hardcover, because there are only 4 paperbacks. So you only have to make sure you get at least one paperback. There are $\binom{10}5$ ways to choose 5 of the books, but we have to subtract the number of "bad" choices, the ones with no paperback. These are the sets of 5 books chosen entirely from the 6 hardcovers, so there are $\binom65$ of them. So the number of good choices is $\binom{10}5-\binom65$.

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