anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
any order 10!
anonymous
  • anonymous
oops i had slow and fast backwards.
anonymous
  • anonymous
huh!

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More answers

anonymous
  • anonymous
i screwed up on #2 not i have to think.
anonymous
  • anonymous
zarkon?
anonymous
  • anonymous
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
anonymous
  • anonymous
How can the first song be both fast and slow?
anonymous
  • anonymous
first two slow 3*2 then 8*7*6*5*4*3
Zarkon
  • Zarkon
is there a reason you are dropping the 2?
Zarkon
  • Zarkon
\[10!\] \[_7P_2\cdot 8!\] \[_3P_2\cdot 8!\]
anonymous
  • anonymous
i have no clue what yall are doing please explain!
Zarkon
  • Zarkon
\[_7P_2=7\times 6\] \[_3P_2=3\times 2\]
Zarkon
  • Zarkon
1) 3628800 2) 1693440 3) 241920
anonymous
  • anonymous
ok so it was right yes? i mean what i wrote
anonymous
  • anonymous
oh no it was wrong. i didn't put the 2 at the end. doh
anonymous
  • anonymous
first two slow 3*2 then 8*7*6*5*4*3*2 so 3*2*8*7*6*5*4*3*2
anonymous
  • anonymous
i would like to think that was a typo
Zarkon
  • Zarkon
I would too
anonymous
  • anonymous
lol!

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