Approach

If every problem had a distinct label, there would be ~n!~ ordered problemsets.

Problems with the same label are indistinguishable, so this overcounts. For each label ~i~ the ~a_i~ copies can be permuted freely without changing the sequence, so divide by ~a_i!~.

The number of distinct sequences is the multinomial coefficient

~\dfrac{n!}{a_1! \, a_2! \, \cdots \, a_k!}~.

Two useful special cases:

• every ~a_i = 1~ (so ~k = n~): the labels are all distinct and the answer is ~n!~;
• one ~a_i = n~ and the rest ~0~: every problem has the same label (or unused labels contribute ~0! = 1~) and the answer is ~1~.

Another checkpoint: with two labels the formula is ~\binom{n}{a_1} = n! / (a_1! \, a_2!)~.

Precompute factorials and inverse factorials modulo ~10^9 + 7~ up to ~n~, then multiply ~n! \cdot \prod (a_i!)^{-1}~. Time is ~O(n)~.

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())
a = list(map(int, input().split()))

ans = fact[n]
for x in a:
    ans = ans * invfact[x] % MOD
print(ans)