Approach

A presenting sequence is an ordered selection of ~k~ distinct people out of ~n~. That is the falling factorial (a permutation)

~P(n, k) = n! / (n - k)!~

when ~0 \le k \le n~. If ~k > n~ there are not enough people, so the answer is ~0~.

This is not a combination ~\binom{n}{k}~: swapping two chosen people produces a different speaking order, so two sequences that use the same set of people still count separately.

The empty product (~k = 0~) is ~1~, including the case ~n = k = 0~.

Precompute factorials and inverse factorials modulo ~10^9 + 7~ up to ~10^6~, then answer in constant time as ~n! \cdot (n - k)!^{-1}~. Time is ~O(10^6)~ for the precomputation.

Solution (Python)

MOD = 10**9 + 7
MAXN = 10**6 + 5

fact = [1] * MAXN
for i in range(1, MAXN):
    fact[i] = fact[i - 1] * i % MOD
invfact = [1] * MAXN
invfact[MAXN - 1] = pow(fact[MAXN - 1], MOD - 2, MOD)
for i in range(MAXN - 2, -1, -1):
    invfact[i] = invfact[i + 1] * (i + 1) % MOD

n, k = map(int, input().split())
if k > n:
    print(0)
else:
    print(fact[n] * invfact[n - k] % MOD)