I am studying for my GED. I was homeschooled, so most of the math topics covered are fairly easy, but there's one which I never went over, and which is giving me some trouble. I understand how to do the problems. The math works out. The trouble is that I don't understand why the process to get the solution is different. I was hoping someone here could explain it in terms I can understand.

I'm dealing with what my study guide terms 'counting' questions. They deal with the number of possible choices in a given scenario. Here's a typical question:

A DJ has enough time to play four songs. She has seven different songs to choose from. How many different orderings of songs can she choose?

This is fairly straightforward. The work looks like $(7)(6)(5)(4)$. I understand why this works. She has seven choices first, then six, and so on. It's the next type of question which I can't figure out:

A DJ is choosing four new records for his collection. He has seven available choices. How many different groups of records can he choose?

This one starts out like the first one, but there's an additional step. You divide thusly: $\frac{(7)(6)(5)(4)}{(4)(3)(2)(1)}$. I know where the numbers are coming from, and how the math works. But I would like to know why. If I understand the principle, the problem will be much easier.

The book explains that this is because 'order does not matter'. That doesn't make any sense, as clearly order doesn't matter in the first problem either. There's no further explanation.

I need to understand why there is an additional step, so that I will be able to spot the differences between the two problems on the test.

Note: I'm not asking for how to solve the problem, or how the math works. I'm asking why two problems, which, to me, look identical, are solved differently.

  • $\begingroup$ Technically these are known as Permutations (the first question) and Combinations (the second). $\endgroup$ – JefferMC Mar 8 '18 at 3:44
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    $\begingroup$ Ii is clearly stated in the first example that the order does matter: 'How many different orderings of songs can she choose?' $\endgroup$ – CiaPan Mar 8 '18 at 10:16
  • $\begingroup$ I wonder if it's reasonable to assume that the DJ won't play the same song four times in a row. I've done that before at pub jukeboxes.... $\endgroup$ – mcintyre321 Mar 8 '18 at 11:42
  • $\begingroup$ @mcintyre321 One of the many reasons I dislike word problems... too much room for interpretation. $\endgroup$ – Thomas Reinstate Monica Myron Mar 8 '18 at 15:18

Ordering does matter in the first question. It even says so - "how many different orderings of songs can she choose?". So in the first one, playing "Twinkle Twinkle Little Star" before "Happy Birthday To You" is a different result to playing "Happy Birthday" first.

In the second one, the DJ is choosing groups of records, i.e. he's just pulling them out in any order. So in that case, the set {"Toto I", "Toto II", "Toto III", "Toto IV"} is considered the same set as {"Toto II", "Toto IV", "Toto I", "Toto III"}.

To switch between the two, you can start with the first case (pick 4 things, in order, from a set of 7), then having listed all of those sets, you note that they can be grouped together into equal-sized sets of redundant orderings. For example, let's say we're just picking 2 options out of {A, B, C, D}. So if order does matter, then the options are:


But you can then group those together into options that are re-orderings of each other, i.e.


So you picked $4 \times 3$, but then grouped them into sets of $2 \times 1$, giving you $\frac{(4)(3)}{(2)(1)}$ different options in the end.

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    $\begingroup$ Which is why the denominator is exactly the number of possible orderings of the items chosen. $\endgroup$ – hobbs Mar 8 '18 at 7:29
  • $\begingroup$ I have to admit I still can't pick out which problem is which when presented with several options. Would you be willing to explain this further in chat? $\endgroup$ – Thomas Reinstate Monica Myron Jun 14 '18 at 23:12
  • $\begingroup$ Probably not, but generally you want to look for clues that answer the question - if I rearrange the items I pick, does that count as the same selection? So if I'm just choosing what things to pack in a bag then it doesn't matter which way around they get picked, but if I'm choosing what to wear on 3 consecutive days it probably does. $\endgroup$ – ConMan Jun 14 '18 at 23:23

Let's compare the two question:

Question $1$:

A DJ is choosing four new records for his collection. He has seven available choices. How many different orderings of songs can she choose

Quesiton $2$:

A DJ is choosing four new records for his collection. He has seven available choices. How many different groups of records can he choose?

The first question is like someone asking the DJ how many possible ordering is there? He cares about the ordering of the songs being played. tell me exactly what is the first song, the second song, the third and the fourth. How many possible answers are there.

The second question is someone just asking what are the songs being played tonight. We just want to know what are the songs, I don't care about the order the songs that they will be played.

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The order does matter in the first problem: for the first song, you have $7$ choices, after which you have $6$ choices left for the second song, etc. That is, you are creating an ordered playlist in the first problem.

For the second problem, you are asked to create a 'collection' of songs, and for a collection, order does not matter.

So yes, these two problems are different.

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