A disc jockey has 10 songs to play. Seven are slow songs, and three are fast songs. Each song is to be played only once. In how many ways can the disc jockey play the 10 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.
a) -- 10 songs played in any order = 10!
b) -- first song is slow, and last song is slow
no. of ways = 7x8x7x6x5x4x3x2x1x6 = 42(8!)
c) frist 2 must be fast:
no. of ways = 3x2x8! = 6(8!)
To determine the number of ways the disc jockey can play the 10 songs, we can approach each scenario separately.
1. The songs can be played in any order:
In this case, we are looking for the number of permutations of all 10 songs. Since each song can only be played once, we can use the factorial function. The total number of ways the disc jockey can play the songs in any order is given by:
10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3,628,800 ways.
2. The first song must be a slow song and the last song must be a slow song:
Here, we need to arrange the 8 remaining songs (7 slow songs and 1 fast song), since the first and last songs are already determined. Again, we can use the factorial function to calculate the number of arrangements:
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320 ways.
3. The first two songs must be fast songs:
In this case, the first two songs are fixed. We need to arrange the remaining 8 songs (5 slow songs and 3 fast songs). The number of arrangements is given by:
8! = 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 40,320 ways.
Therefore, the number of ways the disc jockey can play the 10 songs, considering the given conditions, is 40,320.