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DJ Howard is making a playlist for a friend; he is trying to decide what 9 songs to play and in what order they should be played. If he has his choices narrowed down to 7 pop, 3 hip-hop, 6 country, and 7 blues songs, and he wants to play no more than 3 country songs, how many different playlists are possible? Express your answer in scientific notation rounding to the hundredths place.

Respuesta :

Answer:

1.87 × 10¹⁰ possibilities

Step-by-step explanation:

For Possibility which order/arrangement is significant, we use the Permutation method:

[tex]\boxed{_nP_r=\frac{n!}{(n-r)!} }[/tex]

where,

  • P = total number of outcomes (possibilities)
  • n = total number of objects
  • r = number of selected objects

Since there is a restriction for Country songs, then we group Country songs in 1 group and all other songs into non-Country group.

  • Total number of Country songs = 6
  • Total number of non-Country songs = 7+3+7 = 17

For Possibility which no more than 3 Country songs, there is 4 events where:

  1. 0 out of 6 Country songs + 9 out of 17 non-Country songs
  2. 1 out of 6 Country songs + 8 out of 17 non-Country songs
  3. 2 out of 6 Country songs + 7 out of 17 non-Country songs
  4. 3 out of 6 Country songs + 6 out of 17 non-Country songs

Event 1:

[tex]Outcome_1=\ _6P_0\times_{17}P_9[/tex]

                [tex]\displaystyle=\frac{6!}{(6-0)!} \times\frac{17!}{(17-9)!}[/tex]

                [tex]\displaystyle=\frac{6!}{6!} \times\frac{17!}{8!}[/tex]

                [tex]=1\times(17\times16\times15\times...\times9)[/tex]

                [tex]=8,821,612,800[/tex]

Event 2:

[tex]Outcome_2=\ _6P_1\times_{17}P_8[/tex]

                [tex]\displaystyle=\frac{6!}{(6-1)!} \times\frac{17!}{(17-8)!}[/tex]

                [tex]\displaystyle=\frac{6!}{5!} \times\frac{17!}{9!}[/tex]

                [tex]=6\times(17\times16\times15\times...\times10)[/tex]

                [tex]=5,881,075,200[/tex]

Event 3:

[tex]Outcome_3=\ _6P_2\times_{17}P_7[/tex]

                [tex]\displaystyle=\frac{6!}{(6-2)!} \times\frac{17!}{(17-7)!}[/tex]

                [tex]\displaystyle=\frac{6!}{4!} \times\frac{17!}{10!}[/tex]

                [tex]=(6\times5)\times(17\times16\times15\times...\times11)[/tex]

                [tex]=2,940,537,600[/tex]

Event 4:

[tex]Outcome_4=\ _6P_3\times_{17}P_6[/tex]

                [tex]\displaystyle=\frac{6!}{(6-3)!} \times\frac{17!}{(17-6)!}[/tex]

                [tex]\displaystyle=\frac{6!}{3!} \times\frac{17!}{11!}[/tex]

                [tex]=(6\times5\times4)\times(17\times16\times15\times...\times12)[/tex]

                [tex]=1,069,286,400[/tex]

[tex]Total\ outcomes=outcome_1+outcome_2+outcome_3+outcome_4[/tex]

                         [tex]=259,459,200+5,881,075,200+2,940,537,600+1,069,286,400[/tex]

                         [tex]=18,712,512,000[/tex]

                         [tex]=1.87\times10^{10}[/tex]