Combinatoric 'N choose R' in java math?

Is there a built in method in a java library that can compute 'N choose R' for any N, R?

Solution 1:

The Formula

It's actually very easy to compute N choose K without even computing factorials.

We know that the formula for (N choose K) is:

    N!
 --------
 (N-K)!K!

Therefore, the formula for (N choose K+1) is:

       N!                N!                   N!               N!      (N-K)
---------------- = --------------- = -------------------- = -------- x -----
(N-(K+1))!(K+1)!   (N-K-1)! (K+1)!   (N-K)!/(N-K) K!(K+1)   (N-K)!K!   (K+1)

That is:

(N choose K+1) = (N choose K) * (N-K)/(K+1)

We also know that (N choose 0) is:

 N!
---- = 1
N!0!

So this gives us an easy starting point, and using the formula above, we can find (N choose K) for any K > 0 with K multiplications and K divisions.

Easy Pascal's Triangle

Putting the above together, we can easily generate Pascal's triangle as follows:

    for (int n = 0; n < 10; n++) {
        int nCk = 1;
        for (int k = 0; k <= n; k++) {
            System.out.print(nCk + " ");
            nCk = nCk * (n-k) / (k+1);
        }
        System.out.println();
    }

This prints:

1 
1 1 
1 2 1 
1 3 3 1 
1 4 6 4 1 
1 5 10 10 5 1 
1 6 15 20 15 6 1 
1 7 21 35 35 21 7 1 
1 8 28 56 70 56 28 8 1 
1 9 36 84 126 126 84 36 9 1 

BigInteger version

Applying the formula for BigInteger is straightforward:

static BigInteger binomial(final int N, final int K) {
    BigInteger ret = BigInteger.ONE;
    for (int k = 0; k < K; k++) {
        ret = ret.multiply(BigInteger.valueOf(N-k))
                 .divide(BigInteger.valueOf(k+1));
    }
    return ret;
}

//...
System.out.println(binomial(133, 71));
// prints "555687036928510235891585199545206017600"

According to Google, 133 choose 71 = 5.55687037 × 10³⁸.

References

• Wikipedia/Binomial coefficient
• Wikipedia/Pascal's triangle
• Wikipedia/Combination

Solution 2:

The apache-commons "Math" supports this in org.apache.commons.math4.util.CombinatoricsUtils

Solution 3:

The recursive definition gives you a pretty simple choose function which will work fine for small values. If you're planning on running this method a lot, or on large values, it would pay to memoize it, but otherwise works just fine.

public static long choose(long total, long choose){
    if(total < choose)
        return 0;
    if(choose == 0 || choose == total)
        return 1;
    return choose(total-1,choose-1)+choose(total-1,choose);
}

Improving the runtime of this function is left as an exercise for the reader :)

Solution 4:

I am just trying to calculate number of 2 card combinations with different deck sizes...

No need to import an external library - from the definition of combination, with n cards that would be n*(n-1)/2

Bonus question: This same formula calculates the sum of the first n-1 integers - do you see why they're the same? :)