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.