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keebler01

  • 3 years ago

Show all work. 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.

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  1. satellite73
    • 3 years ago
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    any order 10!

  2. satellite73
    • 3 years ago
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    oops i had slow and fast backwards.

  3. keebler01
    • 3 years ago
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    huh!

  4. satellite73
    • 3 years ago
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    i screwed up on #2 not i have to think.

  5. satellite73
    • 3 years ago
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    zarkon?

  6. satellite73
    • 3 years ago
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    7 for the first, 6 for the last, and 8*7*6*5*4*3 for the middle so i think it is 7*6/8*7*6*5*4*3

  7. xChessMachinex
    • 3 years ago
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    How can the first song be both fast and slow?

  8. satellite73
    • 3 years ago
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    first two slow 3*2 then 8*7*6*5*4*3

  9. Zarkon
    • 3 years ago
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    is there a reason you are dropping the 2?

  10. Zarkon
    • 3 years ago
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    \[10!\] \[_7P_2\cdot 8!\] \[_3P_2\cdot 8!\]

  11. keebler01
    • 3 years ago
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    i have no clue what yall are doing please explain!

  12. Zarkon
    • 3 years ago
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    \[_7P_2=7\times 6\] \[_3P_2=3\times 2\]

  13. Zarkon
    • 3 years ago
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    1) 3628800 2) 1693440 3) 241920

  14. satellite73
    • 3 years ago
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    ok so it was right yes? i mean what i wrote

  15. satellite73
    • 3 years ago
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    oh no it was wrong. i didn't put the 2 at the end. doh

  16. satellite73
    • 3 years ago
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    first two slow 3*2 then 8*7*6*5*4*3*2 so 3*2*8*7*6*5*4*3*2

  17. satellite73
    • 3 years ago
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    i would like to think that was a typo

  18. Zarkon
    • 3 years ago
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    I would too

  19. satellite73
    • 3 years ago
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    lol!

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