let the number of songs be n
C(n,10) > 10 000 000
n!/(10!(n-10)) > 10 000 000
By trial and error, using my C(n,r) key on my calculator
C(25,10) = 3 268 760
C(26,10) = 5 311 735
C(27,10) = 8 436 285
C(28,10) = 13 123 110
1. So they must have at least 28 songs, (not very impressive)
2. Combinations of 5 from those 28
= C(28,5) = 98280
3. prob --- we need 5 of your favourite 5 plus 5 of the remaining 23
prob = C(5,5) x C(23,5)/C(28,10) = .002564
You're friend tells you about a new online music site called ComboAl-
bum that lets you choose any 10 songs from their song library for 9:99. ComboAlbum's advertising
says their song catalogue is so large there are over 10; 000; 000 diff�erent song combinations available
for download.
1.What is the minimum size of their song catalog if this claim is true?
2.If the song catalogue had the number of songs found in the previous part, but you
could download combinations of 5 songs for 4:99, how many di�fferent combinations of 5 songs would
there be?
3.If for 2:99 ComboAlbum will sell you a randomly selected combination of 10 songs,
what is the probability that your favorite 5 songs in the song library are among the 10 selected.
Again, use the total number of songs found in the �rest part.
2 answers
Thank you!