Trick for modular division ( (x1 * x2 .... xn) / b ) mod (m)

Given integers x1, x2, x3......xn, b, and m, we are supposed to find the result of ((x1*x2....xn)/b)mod(m). 
Example 1: Suppose that we are required to find (55C5)%(1000000007) i.e ((55*54*53*52*51)/120)%1000000007 
Naive Method : 

1. Simply calculate the product (55*54*53*52*51)= say x,
2. Divide x by 120 and then take its modulus with 1000000007

Using Modular Multiplicative Inverse : 
The above method will work only when x1, x2, x3....xn have small values. 
Suppose we are required to find the result where x1, x2, ....xn fall in the range of ~1000000(10^6). So we will have to exploit the rule of modular mathematics which says : 
(a*b)mod(m)=(a(mod(m))*b(mod(m)))mod(m)
Note that the above formula is valid for modular multiplication. A similar formula for division does not exist. 
i.e (a/b)mod(m) != a(mod(m))/b(mod(m)) 

1. So we are required to find out the modular multiplicative inverse of b say i and then multiply 'i' with a .
2. After this we will have to take the modulus of the result obtained. 
   i.e ((x1*x2....xn)/b)mod(m)=((x1*x2....xn)*i)mod(m)= ((x1)mod(m) * (x2)mod(m) *.... (xn)mod(m) * (i)mod(m))mod(m)

Note: To find modular multiplicative inverse we can use the Extended Euclidean algorithm or Fermat’s Little Theorem. 
Example 2 : Let us suppose that we have to find (55555C5)%(1000000007) i.e ((55555*55554*55553*55552*55551)/120)%1000000007. 
 

Different Languages give different result for the naive approach. 

C++
Input : 
Output:
 Answer using naive method: 18446744073703577963
 Answer using multiplicative modular inverse concept: 125376140
 
Python 
Input:
Output:
 Answer using naive method: 300820513
 Answer using multiplicative modular inverse concept: 125376140
 
JavaScript:
Input:
Output:
 Answer using naive method: 301201761
 Answer using multiplicative modular inverse concept: 125376140

It is clear from the above example that the naive method will lead to an overflow of data resulting in an incorrect answer. Moreover, using modular inverse will give us the correct answer.
Without Using Modular Multiplicative Inverse : 
But it is interesting to note that a slight change in code will discard the use of finding modular multiplicative inverse. 
 

Answer using shortcut: 300820513

Why did it work? 
This will work only in case when the denominator is a factor of numerator i.e. when a % b = 0 following the rule: 
If b | a, then we can write (a/b) % p as (a % p*b)/b. 
This rule proves useful for small values of b.
Let us consider a = x1*x2*x3.......xn 
We have to find ans = (a/b)%1000000007 

1. Let result of a%(1000000007*b) be y. To avoid overflow, we use modular multiplicative property. This can be represented as 
   a = (1000000007*b)q + y where y < (1000000007*b) and q is an integer
2. Now dividing LHS and RHS by b, we get 
   y/b = a/b -(1000000007*b)*q/b 
   = a/b -1000000007*q < 1000000007 (From 1) 
   Therefore, y/b is equivalent to (a/b)mod(b*1000000007). :)