Question 363845: Hello Can someone please check my answer for me. Thank you so much.
A disc jockey has 8 songs to play. Six are slow songs, and two are fast songs. Each song is to be played only once. In how many ways can the disc jockey play the 8 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.
Answer:
a).The songs can be played in any order slow, slow, slow, fast, slow, fast, slow, slow. = 3slow+ 1 fast; + 3 slow +1 fast = 8
b). The first song must be a slow song and the last song must be a slow
1 slow, 1fast, 1slow, 1fast, 1slow,1slow, 1slow, 1slow = 2/8 or 1/4
c) The first two songs must be fast songs.
Fast, fast, slow, slow, slow, slow, slow, slow
=2/8 or 1/4
Thank you,
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! A disc jockey has 8 songs to play. Six are slow songs, and two are fast songs. Each song is to be played only once. In how many ways can the disc jockey play the 8 songs if
• The songs can be played in any order.
The 1st song is 1 of 8, then 1 of 7, then 6, etc
= 8*7*6*5... = 8!
= 40320
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• The first song must be a slow song and the last song must be a slow song.
The 1st is 1 of 6, the last is 1 of 5, the others are 1 of 6,5,4,3,2,1
= 30*6! = 21600
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• The first two songs must be fast songs.
There are only 2 fast songs, so that's 2*1
The others are 6*5*4*3*2*1 = 6!
--> 1440
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