- #1
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A friend asked me this question, and I am not sure I know how to answer it:
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I have 15 items, and I want to figure out how many unordered combinations of three there are. Any ideas?
a b c d e f g h i j k l m n o
Example:
abc bcd cde
abd bce cdf
abe bcf cde
abf bcg cdg
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The way I think it can be done is using this formula:
[tex]
{n\choose k}={n!\over k!(n-k)!}
[/tex]
so you would have [tex]{15!\over 3!12!}[/tex] or 455 possible combinations.
But I am not really confident in this because I never studied this formula in a class - and I don't know if I am using it correctly.
Thanks!
-MIH
************************************************
I have 15 items, and I want to figure out how many unordered combinations of three there are. Any ideas?
a b c d e f g h i j k l m n o
Example:
abc bcd cde
abd bce cdf
abe bcf cde
abf bcg cdg
************************************************
The way I think it can be done is using this formula:
[tex]
{n\choose k}={n!\over k!(n-k)!}
[/tex]
so you would have [tex]{15!\over 3!12!}[/tex] or 455 possible combinations.
But I am not really confident in this because I never studied this formula in a class - and I don't know if I am using it correctly.
Thanks!
-MIH