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You have 10 songs to burn onto a CD. how many different ways can the CD be made?

I need to know how many ways the CD can be made

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Sky W. | Patient and Knowledgeable Math TutorPatient and Knowledgeable Math Tutor
5.0 5.0 (12 lesson ratings) (12)
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Think about it like positions of a sports team. You have 10 players that you need to rearrange for your roster. You can repeat any of the players (or songs). So for position 1 (song 1) you have 10 choices because you haven't picked any to use yet. For spot 2, you already used one player (1 song) so you have 9 left to choose from. Keep going to slot 3, there are 8 remaining, then 7, 6, 5, 4, 3, 2, 1. this is also called FACTORIAL, or written at 10!. To find the answer you would multiply 10*9*8*7...*2*1.  The reason you can do this is because you don't have any restrictions on what song goes in what place, but you also don't want to repeat any of the songs.
 
I hope that helped. Any other questions feel free to ask.
David S. | Wise Math TutorWise Math Tutor
5.0 5.0 (57 lesson ratings) (57)
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Brittany...This would be a permutation problem. 10 things taken 10 at a time.
 
The answer would be 10! That is 10 factorial which is 10*9*8*7*6*5*4*3*2*1 = 3,628,800

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In general a permutation is n!/(n-r)! where n is the number of things you have and r is how many you are taking at a time.  0! is agreed to = 1
 
So for our problem above we have 10!/(10-10)! which is 10!/1 = 3,628,800
Arthur D. | Effective Mathematics TutorEffective Mathematics Tutor
5.0 5.0 (6 lesson ratings) (6)
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10*9*8*7*6*5*4*3*2*1=3,628,800
this is called a permutation, an arrangement of objects(songs) in a specific order
for example, if you have songs A,B, and C you have:
ABC,ACB,BAC,BCA,CAB, and CBA
3 factorial =3 !=3*2*1=6 permutations
Vivian L. | Microsoft Word/Excel/Outlook, essay composition, math; I LOVE TO TEACHMicrosoft Word/Excel/Outlook, essay comp...
3.0 3.0 (1 lesson ratings) (1)
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Hi Again Brittany;
Another great question...
The equation for this would be...
10!
10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1
3628800 possible arrangements.
This is because for the 
first selection, there are 10 to choose from...
second selection, there are 9 to choose from...
third selection, there are 8 to choose from...
etc.