How many ways can a disc jockey play 11 songs

aburchett

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A disc jockey has 11 songs to play. Seven are slow songs, and four are fast songs. Each song is to be played only once. In how many ways can the disc jockey play the 11 songs if
a) The songs can be played in any order.
b) The first song must be a slow song and the last song must be a slow song.
c) The first two songs must be fast songs.

My work:
a) 11!= 11*10*9*8*7*6*5*4*3*2*1= 39,916,800

b) 11_P_4= 11!/4!= 11*10*9*8*7*6*5*4*3*2*1/4*3*2*1= 1,663,200

c) 11_P_7= 11!/7!= 11*10*9*8*7*6*5*4*3*2*1/7*6*5*4*3*2*1= 7,920

Is this correct?
 
aburchett said:
A disc jockey has 11 songs to play. Seven are slow songs, and four are fast songs. Each song is to be played only once. In how many ways can the disc jockey play the 11 songs if
a) The songs can be played in any order.
b) The first song must be a slow song and the last song must be a slow song.

You can choose two slow songs (first and last) out of seven in 7_P_2 ways

Now you have 9 songs that can be played in any order - 9! ways

So total ways = (7_P_2) * (9!)


c) The first two songs must be fast songs.

Follow the method above.

My work:
a) 11!= 11*10*9*8*7*6*5*4*3*2*1= 39,916,800

b) 11_P_4= 11!/4!= 11*10*9*8*7*6*5*4*3*2*1/4*3*2*1= 1,663,200

c) 11_P_7= 11!/7!= 11*10*9*8*7*6*5*4*3*2*1/7*6*5*4*3*2*1= 7,920

Is this correct?
 
I'm a little confused how to calculate this:

(7_P_2) * (9!) =
7!/2! * 9!=
(7*6*5*4*3*2*1/2*1) * 9*8*7*6*5*4*3*2*1=
2520(362880)=
914,457,600

Is this how?
 
Or should it be set up like this:

P[sub:3mi3eezo]7,2[/sub:3mi3eezo]=
7!/(7-2)!=
7!/5!=
7*6*5*4*3*2*1/5*4*3*2*1=
42
 
aburchett said:
I'm a little confused how to calculate this:

(7_P_2) * (9!) =
7!/2! * 9!=

7_P_2 = 7!/(7-2)! = 42

(7*6*5*4*3*2*1/2*1) * 9*8*7*6*5*4*3*2*1=
2520(362880)=
914,457,600

Is this how?
 
There wasn't any message with the last post. Is the post you picked the correct way to determine this problem?
 
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