If the number of non negative integer solutions for the equation \(x_1 + x_2 \space + \space ..+ \space x_n \space = \space n\), then the number of ways the distribution can be done is = \(^{n+r−1}C_{r−1}\). 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. \(^{n+r−1}C_{r−1}= ^{15}C_2 = 105\).

When Ram has 1 chocolates, then S' + B + G = 12. n = 12 and r = 3. \(^{n+r−1}C_{r−1}= ^{14}C_2 = 91\).

When Ram has 2 chocolates, then S' + B + G = 11. n = 12 and r = 3. \(^{n+r−1}C_{r−1}= ^{13}C_2 = 78\).

When Ram has 3 chocolates, then S' + B + G = 10. n = 12 and r = 3. \(^{n+r−1}C_{r−1}= ^{12}C_2 = 66\).

When Ram has 4 chocolates, then S' + B + G = 9. n = 12 and r = 3. \(^{n+r−1}C_{r−1}= ^{11}C_2 = 55\).

When Ram has 5 chocolates, then S' + B + G = 8. n = 12 and r = 3. \(^{n+r−1}C_{r−1}= ^{10}C_2 = 45\).


Total number of ways = 105 + 91 + 78 + 66 + 55 + 45 = 440


Option E

Arun Kumar
_________________