Evaluate the binomial theorem!

Question

Your mission, even if you don't accept it, is to input three numbers from STDIN or file:

5 0.5 6

What you must do is evaluate this:

(5a + 0.5b)^6

This basically means that the first two numbers are the coefficients of a and b and the third is the exponent in a binomial expansion.

Rules:

• The first two numbers are either integers or simple terminating fractions.
• The third number is a positive integer.
• Do not use any inbuilt binomial functions from libs. (thanks to comments)
• The shortest code wins!

Go on, do it.

• Peter Taylor: The interesting part of this has already been done as Generate Pascal's triangle
• Mr. Llama: What's the objective winning criteria?
• marinus: @PeterTaylor: not necessarily, for example the APL-like languages have binomials built in (as ! at least in APL and J). Indeed, the winner of that question used J's !. It would certainly have been longer if he had to hand-code it, and assuming that this question does not allow their use.
• flornquake: @GigaWatt: I'm assuming shortest code wins, since it's tagged as code-golf.

Answer 1

Mathematica 36

InputForm@Expand[(#1 a + #2 b )^#3] &

Usage

InputForm@Expand[(#1 a + #2 b )^#3] &[5, .5, 6]

Result:

15625*a^6 +9375.*a^5*b +2343.75*a^4*b^2 +312.5*a^3*b^3 +23.4375*a^2*b^4 + 0.9375*a*b^5 + 0.015625*b^6

Answer 2

F# 153 chars 156 bytes

let a,b,n=1.,2.,3
let rec C k=if k=0 then 1 else (n-k+1)*C(k-1)/k
List.iter(fun i->printf"%+ga^%ib^%i"(pown a (n-i)*pown b i*float(C i)) (n-i) i)[0..n]

Answer 3

Python 2, 144 bytes

a,b,n=input();f=lambda n:-~(n and~-n*f(~-n))
for k in range(-~n):print"+ "[0**k]+"%d*(%d*a^%d)*(%d*b^%d)"%(f(n)/f(k)/f(n-k),a**(n-k),n-k,b**k,k)

Try it online!

Answer 4

C (C99), 223 chars

#include <stdio.h>
#include <math.h>
int f(int n,int k){
 return k==0?1:(n*f(n-1,k-1))/k;
}
int main(void){
 float a,b;int n;
 scanf("%f%f%d",&a,&b,&n);
 for(int k=0;k<=n;k++)
 printf("%d*(%fa)^%d*(%fb)^%d+",f(n,k),a,n-k,b,k);
 puts("0");
}

Input:

1 1 2

Output:

1*(1.000000a)^2*(1.000000b)^0+2*(1.000000a)^1*(1.000000b)^1+1*(1.000000a)^0*(1.000000b)^2+0

• Johannes Kuhn: Do you have check if k==0? I think a simple k would do the same.
• Mr. Llama: My brief attempt at optimization: ideone.com/tUlfKs
• md5: @GigaWatt: Well, it wouldn't be C99 then.