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I take great offense in calling my nCr function horribly inaccurate.

You do say right away that is a valid formula for binomial coefficients, but then you say it's hardly a good one?

It is a formula for calculating r-combinations, not binomial coefficients.

The formula for binomial expansion is:

```          n
_
(x+y)^n = \ C(n,j) * x^(n-j) *y^j
/
¯
j=0
```

And btw, it is the definitive formula for calculating r-combinations. It's like saying 4/2 is not the same as 2. While 2 is a better way to write 4/2, not every one recognizes 2 right away, and I feel, for the benefit of the reader of me 'craft', n!/(r!(n-r)! is a better way to go.

As for your function, it is shorter, but a litle mysterious when it comes to the math.

```
9!         9*8*7* 6!      9*8*7   504
nCr(9,3) = ---------   = ----------- = ------ = --- = 84
3!*(9-3)!     3*2*1*(6!)     3*2*1    6

which follows from the fact that nCr(n,r) = nCr(n,n-r)

n!              n!               n!         _    n!
------- = ----------------- = -------------  = ---------
r!(n-r)    (n-r)!(n-(n-r))!   (n-r)!(n-n+r)!   (n-r)!(r)!
```
This however only works for n>r, as long as n and r are both non-negative integers, but shouldn't be a problem in this case.

As for things being integer, since the input is an integer, it follows that any integer multiple of that integer, will also be an integer.

update:
I give, I give, theory v. practice, there is a difference.

___crazyinsomniac_______________________________________
Disclaimer: Don't blame. It came from inside the void

perl -e "\$q=\$_;map({chr unpack qq;H*;,\$_}split(q;;,q*H*));print;\$q/\$q;"

In reply to (crazyinsomniac) Re: (2) Binomial Expansion by crazyinsomniac
in thread Binomial Expansion by crazyinsomniac

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