1、採用C(a, b) = n! / (m! * (n - m)!)，適用範圍為n <= 20

typedef long long ll;
const int maxn=20+5;
ll a[maxn];
void init() {
	a[0]=1;
	for(int i=1; i<maxn) 
		a[i]=a[i-1]*i;
}
ll get_c(ll n,ll m) {
	if(n<m) return 0;
	return  a[n]/a[m]/a[n-m];
}

2、採用拓展歐幾里得（較快）

const int MOD=1e9+9
LL fac[N];
void init()
{
    LL i;
    fac[0]=1;
    for (LL i = 1; i < N; i++)
	fac[i] = fac[i - 1] * i % MOD;
}
LL exgcd(LL a, LL b, LL &x, LL &y) {
    if (!b) {x = 1; y = 0; return a;}
    LL d = exgcd(b, a % b, y, x);
    y -= a / b * x;
    return d;
}
 
LL inv(LL a, LL n) {
    LL x, y;
    exgcd(a, n, x, y);
    return (x + n) % n;
}
 
LL C(LL n, LL m) {
    return fac[n] * inv(fac[m] * fac[n - m] % MOD, MOD) % MOD;
}
 

3、費馬小定理加快速冪（g^(MOD-2)=1/g%MOD,將除法問題轉化為乘法）

LL da[MAXN];//G++ long long
void init()
{
    int i;
    da[0]=1;
    da[1]=1;
    for(i=2;i<MAXN;i++)
        da[i]=i*da[i-1]%MOD;
}
LL quickmod(LL a,LL b)
{
    LL ans=1;
    while(b)
    {
        if(b&1)
        {
            ans=(ans*a)%MOD;
            b--;
        }
        b/=2;
        a=((a%MOD)*(a%MOD))%MOD;
    }
    return ans;
}
LL C(LL a, LL b)
{
    return (da[a]%MOD)*(quickmod(da[b]*da[a-b]%MOD,MOD-2))%MOD;
}