Show all work. A disc jockey has 12 songs to play. Seven are slow songs, and five are fast songs. Each song is to be played only once. In how many ways can the disc jockey play the 12 songs if

The songs can be played in any order.
The first song must be a slow song and the last song must be a slow song.
The first two songs must be fast songs.

Question

Show all work. A disc jockey has 12 songs to play. Seven are slow songs, and five are fast songs.  Each song is to be played only once.  In how many ways can the disc jockey play the 12 songs if

The songs can be played in any order.
    The first song must be a slow song and the last song must be a slow song.
    The first two songs must be fast songs.

anonymous · Answer

A) The songs can be played in any order: To evaluate a "how many ways" problem, imagine you are going to place the items (songs in this case) into positions.  You then multiply the number of possibilities for each position together.  So, for the first song there are 12 possibilities.  Since each song can only be played once, for the second song, there are only 11 choices.  This continues so you get
$$12\times11\times10\times...\times3\times2\times1=479,001,600$$
B) Again, we're going to place the songs into positions, but this times, we're going to select the first and last songs, then we'll distribute the rest in between.  So, there are 7 possibilites for the first song (since it must be a slow song) and then 6 remaining possibilities for the last song (since it also must be a slow song).  That leaves 10 songs to place in the middle.  For that we place them like we did before giving us
$$7\times6\times10\times9\times...\times2\times1=152,409,600$$
C) This is almost identical to (B).  This time, however, we;ll place the first two fast songs then deal with the rest.  This gives us
$$5\times4\times10\times9\times...\times2\times1=72,576,000$$