Question

A disc jockey (DJ) has 7 songs to play. Four are slow songs, and 3 are fast songs. Each song is to be played only once. If the first song must be a slow song and the last song must be a slow song, the DJ can play the 7 songs in different ways. (Type a whole number.)

Answer 1

The DJ can play the 7 songs in 12 different ways by arranging the remaining three slow songs and two fast songs between the first and last slow songs, which are fixed positions.

Step-by-step explanation:

The question involves solving a problem of permutations with constraints. Given that the first song must be a slow one and the last must also be slow, we are left with five positions, where the remaining songs can be played. Since there are 3 remaining slow songs and 3 fast songs available, we need to calculate the number of ways to arrange these within the five positions. The first and last positions are already decided, so we can focus on the middle five.

We have three slots left for the three remaining slow songs and two slots for the fast songs. We can arrange the three slow songs in the remaining three positions in 3! (factorial) ways, which means 3 x 2 x 1 ways. Similarly, the two fast songs can be arranged in the remaining two positions in 2! (factorial) ways, which is 2 x 1 ways. Therefore, the total number of different ways the DJ can play the 7 songs is calculated by multiplying these two values: 3! x 2!, resulting in a total of 6 x 2 = 12 different ways.