quick question...can a combination problem result in a fraction?

Question

parthkohli · Answer

Of course it can!

parthkohli · Answer

No further explanation.

parthkohli · Answer

$\Large \color{MidnightBlue}{\Rightarrow {nCr =}{n! \over (n - r)!r!} }$
Looks completely like a fraction ;P

kinggeorge · Answer

That's what I was thinking. When you're finding the number of ways to do things, all of the common combinatorics equations result in integers. (except probability things)

lgbasallote · Answer

not probability...

parthkohli · Answer

@lgbasallote I just posted that.

lgbasallote · Answer

by fraction i meant 1/2, 3/2, etc

kinggeorge · Answer

Well you can't have 1.5 ways to do something.

parthkohli · Answer

Yeah, if n and r are numbers, then of course that is a fraction.

lgbasallote · Answer

i dont need philosophical answers =_= @ParthKohli lol

parthkohli · Answer

Well, then I say that a combination can be a fraction.

kinggeorge · Answer

$$\frac{n!}{k!(n-k)!}$$is always an integer. So we can hardly say it's a fraction. I might as well go around saying that we should consider 1 a fraction. 

Parts of the equation are fractions, i.e. $\frac{1}{k!(n-k)!}$ is a fraction, but when taken with the $n!$, it shouldn't be considered a fraction

parthkohli · Answer

Oh I see....I wolframmed it and I got all integers.

campbell_st · Answer

combinations are always positive integers... same for permutations...

lgbasallote · Answer

i see thanks guys