題目連結：https://codeforces.com/problemset/problem/1436/C

先在有序序列二分查詢 pos ，統計出按二分路徑分別需要向左和向右查詢的次數
向左查詢即該位置的數比 x 大，向右查詢即該位置數比 x 小
也就是從比 x 大的數中選出 cntr 個，比 x 小的數中選出 cntl 個，
最後再分別乘上排列數即可
階乘和階乘逆元

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<cmath>
#include<stack>
#include<queue>
using namespace std;
typedef long long ll;

const int maxn = 1010;
const int M = 1000000007;

int n,x,pos;
int a[maxn];
ll fac[maxn],ifac[maxn];

ll qsm(ll i,ll po){
	ll res = 1;
	while(po){
		if(po & 1) res = 1ll * res * i % M;
		po >>= 1;
		i = 1ll * i * i % M;
	}
	return res;
}

ll read(){ ll s=0,f=1; char ch=getchar(); while(ch<'0' || ch>'9'){ if(ch=='-') f=-1; ch=getchar(); } while(ch>='0' && ch<='9'){ s=s*10+ch-'0'; ch=getchar(); } return s*f; }

int main(){
	fac[0] = 1ll; int N = 1000;
	for(int i=1;i<=N;++i) fac[i] = 1ll * fac[i-1] * i % M;
	ifac[N] = qsm(fac[N],M-2);
	for(int i=N-1;i>=0;--i){
		ifac[i] = 1ll * ifac[i+1] * (i+1) % M;
	}
	
	n = read(), x = read(), pos = read();// ++pos;

	for(int i=0;i<n;++i) a[i] = i;
	
	int l = 0, r = n;
	
	int cntl = 0, cntr = 0;
	while(l<r){
		int mid=(l+r)/2;

		if(a[mid] <= pos){
			l = mid + 1;
			if(a[mid] != pos) ++cntl;
		} else{
			r = mid;
			++cntr;
		}

	}

	int tl = x - 1, tr = n - x;
	
	if(tr < cntr || tl < cntl){
		printf("0\n");
	} else{
		ll ans = 0;
		ans = 1ll * fac[tr] * ifac[cntr] % M * ifac[tr-cntr] % M * fac[tl] % M * ifac[cntl] % M * ifac[tl-cntl] % M * fac[cntr] % M * fac[cntl] % M * fac[n-1-cntl-cntr] % M;
		printf("%lld\n",ans);
	}
	
	return 0;
}