# Problem

Given non-negative integers $N$, $K$ and a prime number $P$, find the value of:

$$\binom{N}{K} \mod P$$

Sounds like a simple problem right? There are several ways to solve the problem, depending on the constraints of $N$, $K$, and $P$.

**Direct**: If $N$, $K$ are really small, less than $12$, then we can calculate $\binom{N}{K}$ directly using formula $\frac{N!}{K! \times (N-K)!}$ and find the mod $P$ value. Not gonna work when the values are around $1000$.**DP**: If they are slightly bigger but still around $1000$, then we can use dynamic programming to find the result, using the recurrence: $\binom{N}{K} = \binom{N-1}{K} + \binom{N-1}{K-1}$. Not going to work if values are around $10^5$.**Modular Inverse**: If $P$ is a large prime, say $10^9+7$, then using modular inverse and some precalculation we can still calculate $\frac{N!}{K! \times (N-K)!}$. But still, we will need to run $O(N)$ loop to calculate $N!$. Not going to work if $N$ is too large. Also, if $P$ is small then there won’t be any modular inverse since $P$ will not be coprime with denominator anymore.

So what do we do when $N$ and $K$ are large (let’s say around $10^9$) and prime $P$ is small, let’s say $37$. Is it still possible?

Of course. We use **Lucas Theorem**.

Read More

Post Views: 96