Program to count number of possible humble matrices in Python

Suppose we have two values n and m. We have to find number of possible arrangements of humble matrices of order n x m. A matrix is said to be humble when

• It contains each element in range 1 to n x m exactly once
• for any two indices pairs (i1, j1) and (i2, j2), if (i1 + j1) < (i2 + j2), then Mat[i1, j1] < Mat[i2, j2] should hold.

If the answer is too large then return result mod 10^9 + 7.

So, if the input is like n = 2 m = 2, then the output will be 2, because there are two possible matrices -

1	2
3	4

And

1	3
2	4

To solve this, we will follow these steps −

• p := 10^9+7
• result := a list with value 1
• for x in range 2 to 10^6, do
  • temp := last element of result
  • temp :=(temp*x) mod p
  • insert temp at the end of result
• if m > n, then
  • temp := n
  • n := m
  • m := temp
• prod := 1
• for x in range 1 to m, do
  • prod :=(prod * result[x-1]) mod p
  • prod := (prod^2) mod p
• for x in range 0 to n - m, do
  • prod := (prod * result[m-1]) mod p
• return prod

Example

Let us see the following implementation to get better understanding −

p = 10**9+7

def solve(n, m):
   result = [1]
   for x in range(2,10**6+1):
      temp = result[-1]
      temp = (temp*x) % p
      result.append(temp)

   if(m > n):
      temp = n
      n = m
      m = temp
   prod = 1
   for x in range(1,m):
      prod = (prod * result[x-1]) % p
   prod = (prod**2) % p
   for x in range(n-m+1):
      prod = (prod*result[m-1]) % p
   return prod

n = 3
m = 3
print(solve(n, m))

Input

3, 3

Output

24