Question 550762
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In the first case, you simply have 10 different songs and the fact that some are slow and some are fast has no bearing on the outcome.  There are 10 ways to pick the first song, then for each of those 10 ways there are 9 ways to pick the second song...etc.  Altogether, *[tex \Large 10!].


In the second situation, there are 4 ways to choose the first song times 3 ways to choose the last song.  That leaves 2 slow songs and 6 fast songs to spread across the other 8 positions:  *[tex \Large 8!].  Altogether *[tex \Large 12\ \times\ 8!]


Figure the third one the same way.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
My calculator said it, I believe it, that settles it
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