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.
http://www.jiskha.com/display.cgi?id=1311727672
a) played in any order = 12!
b) 1st and last slow, = 7(10!)(6)
.... 7 in the first spot, 6 in the last spot, leaving 10 for the rest
c) 1st two must be fast = 5(4)(10!)
gosh that's so confusing :(
To solve these types of problems, we can use the concept of permutations, which represent the different ways to arrange a set of elements.
1. Songs can be played in any order:
In this case, we have 12 songs, and each song can be chosen independently of others. Hence, we need to find the number of permutations of 12 songs. We can do this by calculating 12 factorial, denoted as 12!.
12! = 12 x 11 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
= 479,001,600
So, there are 479,001,600 different ways to play all the songs without any restrictions.
2. The first song must be a slow song, and the last song must be a slow song:
In this case, we have 7 slow songs and 5 fast songs. To satisfy the given condition, we need to choose the first and last songs from the slow songs.
Since the first song must be a slow song, we have 7 choices here. After selecting the first song, we have 11 songs left, and since the last song must also be a slow song, we have 6 choices for that.
The remaining 10 songs can be played in any order. So, using the concept of permutations, we can calculate the number of ways to arrange the remaining songs, which is 10! (as described above).
Therefore, the total number of ways to play the 12 songs with the given condition is:
7 x 10! = 7 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 17,807,200
So, there are 17,807,200 different ways to play all the songs, ensuring that the first and last songs are slow songs.
3. The first two songs must be fast songs:
In this case, the first two songs must be selected from the 5 fast songs. We have 5 choices for the first song and 4 choices for the second song.
After selecting the first two songs, we have 10 remaining songs that can be played in any order. So, the number of ways to arrange the remaining songs is 10!.
Therefore, the total number of ways to play the 12 songs with the given condition is:
5 x 4 x 10! = 5 x 4 x 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 86,400
So, there are 86,400 different ways to play all the songs, ensuring that the first two songs are fast songs.