# Program to count number of ways we can throw n dices in Python

Suppose we have a number n, number of faces, and a total value, we have to find the number of ways it is possible to throw n dice with faces each to get total. If the answer is very large mod the result by 10**9 + 7.

So, if the input is like n = 2 faces = 6 total = 8, then the output will be 5, as there are 5 ways to make 8 with 2 6-faced dice: (2 and 6), (6 and 2), (3 and 5), (5 and 3), (4 and 4).

To solve this, we will follow these steps −

• m := 10^9 + 7

• dp := a list of size (total + 1) then fill with 0

• for face in range 1 to minimum of faces, update in each step by total + 1, do

• dp[face] := 1

• for i in range 0 to n - 2, do

• for j in range total to 0, decrease by 1, do

• dp[j] := sum of all dp[j - f] for f in range 1 to faces + 1 when j - f >= 1

• return last element of dp mod m

Let us see the following implementation to get better understanding −

## Example

Live Demo

class Solution:
def solve(self, n, faces, total):
m = 10 ** 9 + 7
dp = [0] * (total + 1)

for face in range(1, min(faces, total) + 1):
dp[face] = 1
for i in range(n - 1):
for j in range(total, 0, -1):
dp[j] = sum(dp[j - f] for f in range(1, faces + 1) if j - f >= 1)
return dp[-1] % m
ob = Solution()
n = 2
faces = 6
total = 8
print(ob.solve(n, faces, total))

## Input

2,6,8

## Output

5

Updated on: 07-Oct-2020

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