Just thinking through the problem, I realized it's just a sideways Pascal's Triangle. Let's say our word is "ABCDE".

• A B C
• B C D
• C D E

Let's look at each individual spot and how many paths there are to them. The "C" in the top right and bottom left have only one path to them, and the same can be said of the "B"s and "A". So, we can write the grid like:

• 1 1 1
• 1 C D
• 1 D E

Now, looking at the remaining "C", there are two ways to reach it. You can either come in from above, or come in from the left. There is only one way to come in from the top and one way to come in from the left, we can say that there are two ways to reach it:

• 1 1 1
• 1 2 D
• 1 D F

Now, for the right-most D, we can see that there are two ways to come in from the left, and one way to come in from above:

• 1 1 1
• 1 2 3
• 1 D E

Following a similar process for the last two letters, we can map out the number of paths to each spot in the matrix as:

• 1 1 1
• 1 2 3
• 1 3 6

Giving us a final answer of 6.

This pattern should look familiar. It is actually a Pascal's Triangle in a different orientation. With a much larger matrix, we would get (sorry for terrible formatting):

• 1 1 1 1 1 1
• 1 2 3 4 5 6
• 1 3 6 10 15 21
• 1 4 10 20 35 46
• 1 5 15 35 70 116
• 1 6 21 46 116 232

For a Pascal's Triangle, one can simply use a Combination function to find the value for any point in the triangle.

I haven't solved it yet, but for those completely clueless about where to even start, hopefully this helps.

Prerequisites to solve this problem

1.fast exponentiation

#define ll long long int
#define mod %1000000007
ll fastexpo(ll a, ll n)
{
    ll ret=1;
    ll b=a;
    while(n)
    {
        if(n&1)ret=(ret*b)mod;  //if(n==odd)	
        b=(b*b)mod;		
        n>>=1; 			// n/=2
    }
    return (ll)ret;
}

2.X^(-1) mod(p) can be written as X^(p-2) mod(p)

If anyone want euclid's extended algo; Here we go;

int mmi(int a, int m){
    if(a==1)return 1;
    int e=m;
    int f=a;
    int x= 0;
    int y= 1;
    unsigned long long  u = 1;
    int v = 0;
    int q=0,r=0,c=0,d=0;
    while(f!=1){
        q = e/f; r = e%f; c = x- q*u; d = y - q*v; x = u; y = v; u = c; v = d; e = f; f = r;
    }
    return (u+m)%m;
}

Passes all TC:

#include <bits/stdc++.h>

using namespace std;

const int p=1000000007;
int n, m, t;

long long fmod(long long N) {    if (N<2) return 1; else return (N*fmod(N-1))%p;    }
long long mmi(long long x)  {
    long long result=1, e=p-2;    
    while (e) {
        if (e & 1) result = (result*x)%p;
        x = (x*x)%p;
        e >>= 1;
    }
    return result;
}

int main()
{
    cin >> t;
    while (t--) {
        cin >> n >> m;
        printf("%lld\n", ((fmod(m+n-2)*mmi((fmod(m-1)*fmod(n-1))%p))%p));
    }
    return 0;
}

Can anyone improve my code, Test cases 5,6,7,8 are timed out:

from math import factorial for _ in range(0, int(raw_input())): m, n = map(int, raw_input().split()) p1 = 1 if (m > n): for i in range(m, m+n-1): p1 = p1 * i print (p1/factorial(n-1))%((10**9)+7) else: for i in range(n, m+n-1): p1 = p1 * i print (p1/factorial(m-1))%((10**9)+7)

Great post! I appreciate you for the effort you take to share your knowledge with people. I can find many things that are still unaware. Thanks for you time and knowledge. Great!!! verizon email problem

I have a different view. If I understand well. We have to go from top-left to bottom-right. We have two types of step: R(ight), D(own). (m-1) -> R (n-1) -> D

So just have to count the Permutations of two type of steps (patterns or ways). The problem if I am right hard to calculate. If I am wrong please let me know!

The solution has little to do with elementary facts about combinatorials, and a lot to do with properties of modular arithmetic where the modulus is a prime number. 1000000007 happens to be a prime number. I copied the solution provided by shell87301, below, only translating it to Python3. It was both gratifying and humiliating to see it work after so many tries on my own.

I thought about it this way: there is nothing special about it being a word (like "AWAY"), all we're trying to do is travel from the top left corner of a matrix to the bottom right by only moving right (R) or down (D).

If the matrix has 'm' rows and 'n' columns, then any valid path needs 'm-1' moves down and 'n-1' moves to the right, so 1 simple path could look like this: RRRRR....RRDDDDD....DD with 'n-1' Right moves and then 'm-1' Down moves.

How many valid paths are there? Well, each valid path has a total of (n+m-2) moves that can be reordered in (n+m-2)! ways. However, all the R's are identical so we need to remove all arrangements that are just shifting the R's around. Same argument applies to the D's, so the total number of valid paths is (n+m-2)! / (n-1)! / (m-1)! (this is why Pascal's triangle is also a correct way of thinking about the number of valid paths). To calculate this, follow the instructions in the discussion of the Sherlock and Permutations problem.

I'm confused how it can be gauranteed that (n-1)!(m-1)! is not a multiple of p, such that Fermat's Little Theorem applies?