A disc jockey (DJ) has 9 songs to play. Three are slow songs, and 6 are fast songs. Each song is to be played only once.
a) In how many ways can the DJ play the 9 songs if the songs can be played in any order?
b) In how many ways can the DJ play the 9 songs if the first song must be a slow song and the last song must be a slow song?
c) In how many ways can the DJ play the 9 songs if the first two songs must be fast songs?
a) If the songs can be played in any order, the DJ can play the 9 songs in $\square$ different ways.
(Type a whole number.)
b) If the first song must be a slow song and the last song must be a slow song, the DJ can play the 9 songs in $\square$ different ways.
(Type a whole number.)
c) If the first two songs must be fast songs, the DJ can play the 9 songs in $\square$ different ways.
(Type a whole number.)
Therefore, the DJ can play the 9 songs in \(6 * 5 * 5,040 = \boxed{151,200}\) different ways.
Step 1 :The total number of songs is 9, denoted as \(n\). The number of songs to play is also 9, denoted as \(r\).
Step 2 :The formula for permutations is \(nPr = \frac{n!}{(n - r)!}\).
Step 3 :Substitute \(n = 9\) and \(r = 9\) into the formula, we get \(9P9 = \frac{9!}{(9 - 9)!} = 9!\).
Step 4 :The factorial of 9, denoted as 9!, is the product of all positive integers from 1 to 9, which is 362,880.
Step 5 :\(\boxed{362,880}\) is the number of ways the DJ can play the 9 songs in any order.
Step 6 :If the first song and the last song must be slow songs, there are 3 choices for the first song and 2 choices for the last song.
Step 7 :This leaves 7 songs (1 slow song and 6 fast songs) to be played in any order in the middle.
Step 8 :The formula becomes \(7P7 = \frac{7!}{(7 - 7)!} = 7!\).
Step 9 :The factorial of 7, denoted as 7!, is the product of all positive integers from 1 to 7, which is 5,040.
Step 10 :Therefore, the DJ can play the 9 songs in \(3 * 2 * 5,040 = \boxed{30,240}\) different ways.
Step 11 :If the first two songs must be fast songs, there are 6 choices for the first song and 5 choices for the second song.
Step 12 :This leaves 7 songs (3 slow songs and 4 fast songs) to be played in any order.
Step 13 :The formula becomes \(7P7 = \frac{7!}{(7 - 7)!} = 7!\).
Step 14 :Therefore, the DJ can play the 9 songs in \(6 * 5 * 5,040 = \boxed{151,200}\) different ways.