Which is better way to calculate nCr

Your Recursive Approach is fine but using DP with your approach will reduce the overhead of solving subproblems again.Now since we already have two Conditions-

nCr(n,r) = nCr(n-1,r-1) + nCr(n-1,r);

nCr(n,0)=nCr(n,n)=1;

Now we can easily build a DP solution by storing our subresults in a 2-D array-

int dp[max][max];
//Initialise array elements with zero
int nCr(int n, int r)
{
       if(n==r) return dp[n][r] = 1; //Base Case
       if(r==0) return dp[n][r] = 1; //Base Case
       if(r==1) return dp[n][r] = n;
       if(dp[n][r]) return dp[n][r]; // Using Subproblem Result
       return dp[n][r] = nCr(n-1,r) + nCr(n-1,r-1);
}

Now if you want to further otimise, Getting the prime factorization of the binomial coefficient is probably the most efficient way to calculate it, especially if multiplication is expensive.

The fastest method I know is Vladimir's method. One avoids division all together by decomposing nCr into prime factors. As Vladimir says you can do this pretty efficiently using Eratosthenes sieve.Also,Use Fermat's little theorem to calculate nCr mod MOD(Where MOD is a prime number).

I think your recursive approach should work efficiently with DP. But it will start giving problems once the constraints increase. See http://www.spoj.pl/problems/MARBLES/

Here is the function which i use in online judges and coding contests. So it works quite fast.

long combi(int n,int k)
{
    long ans=1;
    k=k>n-k?n-k:k;
    int j=1;
    for(;j<=k;j++,n--)
    {
        if(n%j==0)
        {
            ans*=n/j;
        }else
        if(ans%j==0)
        {
            ans=ans/j*n;
        }else
        {
            ans=(ans*n)/j;
        }
    }
    return ans;
}

It is an efficient implementation for your Approach #1

Both approaches will save time, but the first one is very prone to integer overflow.

Approach 1:

This approach will generate result in shortest time (in at most n/2 iterations), and the possibility of overflow can be reduced by doing the multiplications carefully:

long long C(int n, int r) {
    if(r > n - r) r = n - r; // because C(n, r) == C(n, n - r)
    long long ans = 1;
    int i;

    for(i = 1; i <= r; i++) {
        ans *= n - r + i;
        ans /= i;
    }

    return ans;
}

This code will start multiplication of the numerator from the smaller end, and as the product of any k consecutive integers is divisible by k!, there will be no divisibility problem. But the possibility of overflow is still there, another useful trick may be dividing n - r + i and i by their GCD before doing the multiplication and division (and still overflow may occur).

Approach 2:

In this approach, you'll be actually building up the Pascal's Triangle. The dynamic approach is much faster than the recursive one (the first one is O(n^2) while the other is exponential). However, you'll need to use O(n^2) memory too.

# define MAX 100 // assuming we need first 100 rows
long long triangle[MAX + 1][MAX + 1];

void makeTriangle() {
    int i, j;

    // initialize the first row
    triangle[0][0] = 1; // C(0, 0) = 1

    for(i = 1; i < MAX; i++) {
        triangle[i][0] = 1; // C(i, 0) = 1
        for(j = 1; j <= i; j++) {
            triangle[i][j] = triangle[i - 1][j - 1] + triangle[i - 1][j];
        }
    }
}

long long C(int n, int r) {
    return triangle[n][r];
}

Then you can look up any C(n, r) in O(1) time.

If you need a particular C(n, r) (i.e. the full triangle is not needed), then the memory consumption can be made O(n) by overwriting the same row of the triangle, top to bottom.

# define MAX 100
long long row[MAX + 1];

int C(int n, int r) {
    int i, j;

    // initialize by the first row
    row[0] = 1; // this is the value of C(0, 0)

    for(i = 1; i <= n; i++) {
        for(j = i; j > 0; j--) {
             // from the recurrence C(n, r) = C(n - 1, r - 1) + C(n - 1, r)
             row[j] += row[j - 1];
        }
    }

    return row[r];
}

The inner loop is started from the end to simplify the calculations. If you start it from index 0, you'll need another variable to store the value being overwritten.