**Concept: The question tests on the area of Modern Math (Counting/Combinatorics)**

**Solution:**

If the number of non-negative integer solutions for the equation x1+x2 + ..+ xn =n,then the number of ways the distribution can be done is = n+r−1Cr−1

In this case, value of any variable can be zero.

let us now look at the conditions.

=>We have that Shyam must have at least 2 chocolates and Ram can have a maximum of 5 chocolates. This means that Ram can have 0, 1, 2, 3, 4 or 5 chocolates.

Let us Give Shyam 2 chocolates, which takes out one condition.

We now have 13 chocolates. let Shyam + 2 = S', where S' can be 0 as it includes 2 chocolates.

When Ram has 0 chocolates, then S' + B + G = 13. n = 13 and r = 3.

n+r−1Cr−1 = 15C2 = 105

When Ram has 1 chocolates, then S' + B + G = 12. n = 12 and r = 3

n+r−1Cr−1=14 C2=91

When Ram has 2 chocolates, then S' + B + G = 11. n = 12 and r = 3.

n+r−1Cr−1=13C2=78

When Ram has 3 chocolates, then S' + B + G = 10. n = 12 and r = 3.

n+r−1Cr−1=12C2=66

When Ram has 4 chocolates, then S' + B + G = 9. n = 12 and r = 3.

n+r−1Cr−1=11C2=55

When Ram has 5 chocolates, then S' + B + G = 8. n = 12 and r = 3.

n+r−1Cr−1=10C2=45

Total number of ways = 105 + 91 + 78 + 66 + 55 + 45 = 440**(Option E)**