# Combinatorics: how many unique albums…

I want to record a set of music albums so that each one is unique. Each album has ten tracks, and I've recorded 4 versions of each track. How many unique albums can I compile so that no two albums has the exact same set of tracks while still having only one version of track 1 one of track 2 etc...my initial assumption was $4\times10!$ which is $$815915283247897734345611269596115894272000000000$$ but this seems a bit too much (although certainly enough for double platinum;)

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Each album must have 10 different songs. For each song, you can choose one of the four versions. A choice out of 4, done 10 times, gives you $4^{10}$ options.

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I assume that an album is unique if it has the same songs, while Andre further assumes that they must be in the same order. Which one of us is right depends on what your definition of 'unique' is. –  Robert Mastragostino Jul 12 '12 at 2:58
Andre, Robert: thank you for this!...yes each album is unique but still has one of four versions of each track. The order is important in the sense that there's only one track 1, one track 2 etc. But one album could just have version B of track one and version A of all the other tracks...so I think that 4 to the power of ten is the answer right? –  Dario Ameni Jul 12 '12 at 3:06
To clarify, my answer says that track 1, track 2, ... track 2, track 1, ... are the same. Andre's original answer says they're different, though he seems to have extended it to include the justification for both cases. –  Robert Mastragostino Jul 12 '12 at 3:14
Could you have "God Save the Queen" (some version) on Track $1$, and "La Marseillaise" on Track $2$ of some album, and "God Save the Queen" on Track $7$, "La Marseillaise" on Track $8$ of some other album? –  André Nicolas Jul 12 '12 at 3:18
Hi Andre: The idea is to still have ten tracks in the same order so the track 'list' would remain the same on each album yet you'd be listening to a different album because at least one track is different...but altering the order is another possibility we might want to explore as well. –  Dario Ameni Jul 12 '12 at 3:41

The $4^{10}$ gives the number of ways to select the versions, without worrying about the order on the album. Call the actual songs Song $1$, Song $2$, and so on. Then the version of Song $1$ can be chosen in $4$ ways. For each of these, the version of Song $2$ can be chosen in $4$ ways, and so on for a total of $4^{10}$.

Now you want to deal with the ordering of the songs on the album, that is, which song will occupy each track. For each selection of versions, you can choose the ordering in $10!$ ways, for a total of $(4^{10})(10!)$.

If the order on the album doesn't matter, or if you will select one version of Song $1$ for track $1$, one version of Song $2$ for track $2$, and so on, then we end up with the just $4^{10}$. But the factorial in your answer makes me think that you are allowing the order of songs to vary, making $(4^{10})(10!)$ the correct answer.

An alternate way of doing the problem is to look one track after the other. The song for track $1$ can be chosen in $10$ ways. For each way, the version can be chosen in $4$ ways. Once we have done this, the song for track $2$ can be chosen in $9$ ways, and then the version in $4$ ways. Continue. We get a total of $$(10)(4)(9)(4)(8)(4)\cdots (2)(4)(1)(4).$$ This can be simplified to $(4^{10})(10!)$.

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Thanks Andre: My comment to Robert's answer clarifies my question. I tried 40factorial because that's the limit of my knowledge at this moment. I'm not a mathematician, but I love math. The same way many non-musicians love music... –  Dario Ameni Jul 12 '12 at 3:10