#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=n-1;i>=a;i--)
#define mp make_pair
#define fi first
#define se second
#define SZ(x) ((int)(x).size())
#define lli long long int
#define pii pair<int,int>
#define vii vector<pii>
#define pb push_back
#define eb emplace_back
#define em emplace
#define all(v)  v.begin(),v.end()
#define mod 1000000007
#define vi vector<int>
#define vl vector<lli>
#define vll vector<pair<lli,lli>>
#define pll pair<lli,lli>
#define vvi vector<vector<int>>
#define vvl vector<vector<lli>>
#define error(args...) { string _s = #args; replace(_s.begin(), _s.end(), ',', ' '); stringstream _ss(_s); istream_iterator<string> _it(_ss); err(_it, args); }
#define tk(args...) take(args);
#define posLSB(X) __builtin_ctz(X)
#define num1bit(X) __builtin_popcount(X)
#define numlead0(X) __builtin_clz(X)
#define umreserve(X) X.reserve(32768); X.max_load_factor(0.25);
void err(istream_iterator<string> it) {} template<typename T, typename... Args> void err(istream_iterator<string> it, T a, Args... args) { cout << *it << " = " << a << "\n"; err(++it, args...);}
void take() {} template<typename T, typename... Args> void take( T & a, Args & ... args ) { cin>>a; take(args...); }
template<class T>
void printvec(vector<T> &a){
	for(T &x:a){
		cout<<x<<" ";
	}
	cout<<endl;
}
template<class T>
void printarr(T a[],lli n){
	for(lli i=0;i<n;i++){
		cout<<a[i]<<" ";
	}
	cout<<endl;
}
struct HASH{
  size_t operator()(const pair<int,int>&x)const{
    return (size_t) x.first * 37U + (size_t) x.second;
  }
};

#define endl "\n"

lli add(lli a, lli b) {
	a += b;
	if (a >= mod) a -= mod;
	return a;
}

lli mul(lli a, lli b) {
	return a * 1ll * b % mod;
}

inline bool setmin(lli &x, lli y) { return (y < x) ? x = y, 1 : 0; }
inline bool setmax(lli &x, lli y) { return (y > x) ? x = y, 1 : 0; }

lli power_mod(lli a,lli x){if(x==0) return 1;lli y=power_mod(a,x/2);lli ans=(y*y)%mod; if(x%2) ans=(ans*a)%mod;return ans;}
lli inv(lli a){return power_mod(a,mod-2);}
lli power(lli a, lli x){ if(x==0) return 1; lli y= power(a,x/2); lli ans=(y*y); if(x%2) ans*=a; return ans;}

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	int n;
	cin>>n;
	vector<int> a(n);
	for(int i=0;i<n;i++)cin>>a[i];
	int p=0,q=0,r=0;
	for(int i=0;i<n;i++){
		if(a[i]==1)p++;
		else if(a[i]==2)q++;
		else r++;
	}
	vector<vector<vector<double>>> dp(n+1,vector<vector<double>>(n+1,vector<double>(n+1,0)));
	// dp[i][j][k] = i/(n)*dp[i-1][j][k];
	// whenever  we access j+1, we are at k-1 similarly for i+1, we are at j-1... thus we should have k,j,i as loop variables from outer to inner loop in order
	//dp[i][j][k]= ((n-i-j-k)/n)*(1+dp[i][j][k])+(i/n)*(dp[i-1][j][k]+1)+(j/n)*(dp[i+1][j-1][k]+1)+k/n*(dp[i][j+1][k-1]+1)
	// upon solving we get the relation that will be detailed in code inside the loops
	for(int k=0;k<=n;k++){
		for(int j=0;j+k<=n;j++){
			for(int i=0;i+j+k<=n;i++){
				if(i+j+k==0)continue;
				double &x=dp[i][j][k];
				x=n*1.0/(i+j+k);
				if(i-1>=0){
					x+=(i*1.0/(i+j+k))*(dp[i-1][j][k]);
				}
				if(j-1>=0){
					x+=(j*1.0/(i+j+k))*(dp[i+1][j-1][k]);
				}
				if(k-1>=0){
					x+=(k*1.0/(i+j+k))*(dp[i][j+1][k-1]);
				}
			}
		}
	}
	cout.precision(16);
	cout<<dp[p][q][r]<<endl;
}


Measured memory_bytes: 218880