DJ Jacqueline is making a playlist for a website; she is trying to decide what 10 songs to play and in what order they should be played. Step 1 of 2 : If she has her choices narrowed down to 6 blues, 3 pop, 6 rock, and 6 jazz songs, and she wants to play no more than 2 rock songs, how many different playlists are possible? Express your answer in scientific notation rounding to the hundredths place.

possible playlists

Step-by-step explanation:

 Since the order of the playlist is essential here then this actually involves permutation.

 The DJ decides what 10 songs to be played and in the order for playing them. Her choice is narrowed to 6 blues, 3 pop, 6 rock and 6 jazz songs. The total songs = 21 and she wants to play no more than 2 rock songs.

  Since it is confirmed that the DJ doesn't want to play more than 2 rock songs, then her playlist will have to accommodate only 0, 1 or 2 blue songs at most.

 We then need to bisect the three different situations and later sum them up.

  From six rock songs, the DJ can only select 0, 1 or 2. Because this scenario involves order, we are going to consider permutation in tackling the issue at hand.

  In the first situation, there is 0 rock song. This implies that there are ten (10) slots for the remaining 15 non - rock songs.

 That is (6 blues + 3 pop + 6 jazz) = 15.

  We can then arrange them in:

   6P0  × 15P10

6P0 = 6!/(6-0)! = 6!/6! = 1

15P0 = 15!/(15-10)! = 15!/5! = 1.08972 × 10^10

 Then (1.08972 × 10^10) × (1) =  1.08972 × 10^10 ways

In the second situation, there is 1 rock song. This then implies that there are nine (9) slots for the other songs that are not rock. They can be arranged in:

    6P1  ×  15P9

6P1 = 6!/(6-1)! = 6!/5! =

 6×5×4×3×2×1/5×4×3×2×1

 = 6

15P9 =  15!/(15-9)! = 15!/6! = 1,816,214,400

Then 1,816,214,400 × 6 = 1.08973×10^10 ways

 In the third situation, there are 2 rock songs. This means that there are eight (8) slots for the other songs that are not rock songs. This can be arranged in:

    6P2  ×  15P8

 6P2  =  6!/(6-2)! = 6!/4!  = 6×5×4×3×2×1/4×3×2×1  = 30

15P8 = 15!/(15-8)!  = 15!/7!  = 259,459,200

Then 259,459,200 × 30 = 7.78378×10^9 ways.

 We then add the outcome of the three different situations/permutations:

 (1.08972 × 10^10) + (1.08973 × 10^10) + (7.78378 × 10^9)  =

 To the hundredth place we will have:

So (6P0 × 15P10) + (6P1 × 15P9) + (6P2 × 15P8) =   possible playlists.

529 and 531

step-by-step explanation: