Find the number of ways to listen to 4 different CDs from a selection of 15 CDs.
I thought it was 15*14*13*12 but it's not
That would be the answer if order mattered, but it does not matter
Let's say you had 4 CDs labeled A,B,C,D there are 4! = 4*3*2*1 = 24 ways to arrange the four letters (eg: ABCD or ACBD) you have to divide the answer you got by 24 to account for the fact that order doesn't matter
aaaah okay okay so 1365
Smaller example Let's say you had 4 CDs (A,B,C,D) and you pick 2 of them. order doesn't matter 4*3 = 12 if order mattered. The full list of choices is AB, BA AC, CA AD, DA BC, CB BD, DB CD, DC notice how the list above is double of what it should be. We've counted everything twice. Which is why we divide by 2! = 2*1 = 2 to get 12/2 = 6. There are 6 ways to pick the two CDs where order doesn't matter
yes it is 1365 an alternative route is to use the combination formula \[\Large C(n,r) = \frac{n!}{r!*(n-r)!}\]
aaaaah okay okay!!
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