#include <vector>
#include <algorithm>
#include <math.h>
#include <stdio.h>
using namespace std;
int main()
{
vector<int>V;
V.push_back(3);
V.push_back(3);
V.push_back(5);
V.push_back(6);
sort(V.begin(),V.end());
int b=1;
while(next_permutation(V.begin(),V.end()))++b;
printf("%d\n",b);
}

B is 1 at first because next_permutation doesn't count sorted order. Hope this help. ;-)

314C - Sereja and Subsequences from Codeforces Round 187 (Div. 1) needs you to do almost the same, except for subsequences must be non-decreasing and you are to output sum of product of all such subsequences. Try reading the editorial (though it did not help me) or accepted codes.
Hope it helps.

oh yeah! I found it. @NiVeR: N<=10^5, and element is not important.

here is brute code:

#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <vector>
#include <set>

#define rep(i,n) for(int i=0,_n=(n);i<_n;++i)
using namespace std;

typedef vector<int> VI;
set<VI> f;
VI a;
int n,s[100],dp[100];

int main(){
    cin>>n;
    rep(i,n) cin>>s[i];
    rep(i,1<<n){
        a.clear();
        rep(j,n) if (i&(1<<j)) a.push_back(s[j]);
        f.insert(a);
    }
    cout<<f.size()<<'\n';
    return 0;
}

But i has a O(N) solution: We have a1a2...,an the sub sequence that have a[i] is f[i] if a[i] appear before f[i]=sum(f[j],j=1..i-1) if a[i] is first appear f[i]=sum(f[j],j=1..i-1)+1 The result is sum(f[j],j=1..n)+1. P/s: sorry, it's wrong.