Why Does (n 0) Equal 1 in Combinatorics?

lalapnt · Oct 22, 2012

Printed equation (5): S(n) = 0(n choose 0) + 6(n choose 1) + 18(n choose 2) + 18(n choose 3) + 6(n choose 4) + 0(n choose 5) + 0(n choose 6) + ..., with (n choose 0) and a red box around it; below, the note: (n choose 0) = 1, (n choose 1) = n, (n choose 2) = n(n-1)/2!, and so on.

in the above pic, (n 0) = 1?? what topic in math is that? i never saw this in my high school math and only saw it in my uni class. i don't know how that is. can someone please help me?

micromass · Oct 22, 2012

They are binomial coefficients. We define

[tex]\binom{n}{k}=\frac{n!}{k! (n-k)!}[/tex]

You have no doubt seen this before in combinatorics, but with other notation. It is a mystery to me why high school textbooks don't use the [itex]\binom{n}{k}[/itex] notation.

http://en.wikipedia.org/wiki/Binomial_coefficient

lalapnt · Oct 22, 2012

super! and of course i know about binomial coefficients. just like this

$Formula for the binomial coefficient: n! divided by k! times (n minus k)!, written as a fraction.$

thank you very much. now I'm not scared anymore! :D

dextercioby · Oct 22, 2012

It comes from the convention 0!=1.

Why Does (n 0) Equal 1 in Combinatorics?

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