You have n dices each one having s sides numbered from 1 to s.
How many outcomes add up to a specified number k?

For example if we roll four normal six-sided dices we
have four outcomes that add up to 5. 

(1, 1, 1, 2)
(1, 1, 2, 1)
(1, 2, 1, 1)
(2, 1, 1, 1)

Difficulty: EASY
Starter code: def outcome(n, s, k):
	

Solutions:
def outcome(n, s, k):
	if n == 1:
		return 1 if 0 < k <= s else 0
	return sum((outcome(n - 1, s, k - j - 1) for j in range(s))) if k > 0 else 0


def outcome(n, s, k):
	return n <= k <= n * s and (n in (1, k) or sum((outcome(n - 1, s, k - x - 1) for x in range(min(s, k - n + 1)))))


def outcome(n, s, k):
	if n * s * k == 0:
		return 0
	dp = [[0] * (k + 1) for i in range(n + 1)]
	for i in range(1, min(s + 1, k + 1)):
		dp[1][i] = 1
	for i in range(2, n + 1):
		for j in range(1, k + 1):
			for l in range(1, min(s + 1, j)):
				dp[i][j] += dp[i - 1][j - l]
	return dp.pop().pop()


from itertools import product

def outcome(n, s, k):
	return sum((1 for roll in product(range(1, s + 1), repeat=n) if sum(roll) == k))


def outcome(n, s, k):
	from itertools import zip_longest
	dice = [1]
	for _ in range(n):
		dice = [sum(x) for x in zip_longest(*([0] * i + dice for i in range(1, s + 1)), fillvalue=0)]
	return dice[k] if k < len(dice) else 0


def outcome(n, s, k):
	return 0 < k <= s if n == 1 else sum((outcome(n - 1, s, k - v) for v in range(1, min(k - (n - 1), s) + 1) if k - v > 0))


from itertools import product

def outcome(n, s, k):
	return sum((sum(x) == k for x in product(range(1, s + 1), repeat=n)))


outcome = o = lambda n, s, k: s >= k > 0 if n == 1 else sum((o(n - 1, s, k - i - 1) for i in range(min(s, k))))