// Copyright 2020 Tsutomu ISHIKAWA
// Author: Tsutomu ISHIKAWA

#include <bits/stdc++.h>
using namespace std;


int l;
int res;
long long a[1010];

long long calcNumOfCombination(int n, int r){
        long num = 1;
        for(int i = 1; i <= r; i++){
            num = num * (n - i + 1) / i;
        }
        return num;
    }

int main() {
    cin >> l;
    cout << calcNumOfCombination(l-1, 11) << endl;
    return 0;
}

/*

2021-03-29 14:18:10.721371

Category: combi

Rating: ?

*/



/*
#pragma GCC optimize("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
*/
#include<bits/stdc++.h>
using namespace std;
//
//
#define fio ios_base::sync_with_stdio(false);cin.tie(NULL);
#define int long long
#define ll long long
#define ld long double
#define gap ' '
#define endl '\n'

/*
void solve() {
	int n; cin>>n; n--; // we have n-1 positions to split, choose 11 
	int up=1,down=1;
	for(int i=1;i<=11;i++) {
		up*=n--;
		down*=i;
		int g=__gcd(up,down);
		up/=g;
		down/=g;
	}
	cout<<up/down<<endl;
}
let try dp approach....
*/
int dp[201][12];

void solve() {
	int n; cin>>n;
	for(int i=1;i<=n;i++) {
		for(int j=0;j<=11;j++) {
			if(j==0) dp[i][j]=1;
			else
			dp[i][j]=dp[i-1][j-1] + dp[i-1][j];
		}
	}
	cout<<dp[n][11]<<endl;
}
void solve(bool testcase) {
	int t;cin>>t;while(t--)solve();
}
int32_t main() {
	fio
	solve();
	return 0;
}