Question

200 identical candies are to be sent to 5 families (A, B, C, D, E). If each family must get at least 3 candies and the family “A” cannot have more than 30 candies, how many different ways are there to distribute the candies?

Answer 1

This is a case of multinomial theorem. Let the candies given to 5 families be a, b, c, d and e

Also A = a - 2,
B = b - 2 and so on .. E = e - 2

If we are given here that:
a, b, c, d, e >= 3

Therefore, A, B, C, D, E >= 1

Therefore, we have here:
a + b + c + d + e = 200
A + B + C + D + E = 200 + 2*5 = 210

Where each one of them is greater than 1.

Therefore the number of solutions of the above equation is computed using the multinomial formula as:

Now let a > 30 candies that is a get more than 30 candies

A + 2 > 30

A > 28 or A >= 29

Let X = A - 28

Therefore, X >= 1

Putting it in the above equation, we have here:
X + 28 + B + C + D + E = 210
X + B + C + D + E = 210 - 28 = 182

with each one of them: X, B, C, D , E >= 1

Number of solutions of above equation using multinomial theorem:

Therefore the number of different ways to distribute the candies here is computed as: