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The stars and bars method is a quick way to calculate the combinations in multi-set problems like this. It is used by transforming the problem into the problem of placing 'N' indistinguishable objects (in this case white balls) into 'B' distinguishable boxes (in this case containers). It is carried out by arranging all objects in a row (stars) and then using 'B-1' bars to divide them into 'B' different boxes.

In this case, we have 8 identical balls (stars) and 4 boxes (bars). However, since no box can be empty, we first place one ball in each box. This leaves us with 4 balls that can be distributed freely. Hence, the problem reduces to the number of ways of distributing 4 balls into 4 boxes which is given by the formula \(C(n + k - 1, k - 1)\), where \(C\) is the binomial coefficient, \(n\) is the number of balls, and \(k\) is the number of boxes. So, the number of ways is \(C(4 + 4 - 1, 4 - 1) = C(7, 3)\).

When the fourth container has an odd number of balls in it, this means that the number of balls in this container could be 1, 3, 5, or 7. We can think of this as four separate cases and calculate the number of ways for each case. Case 1: If the fourth box has 1 ball, we then distribute 7 balls over 4 boxes with no other restrictions, which is \(C(7 + 4 - 1, 4 - 1) = C(10, 3)\) ways.Case 2: If the fourth box has 3 balls, we then distribute 5 balls over 4 boxes with no other restrictions, which is \(C(5 + 4 - 1, 4 - 1) = C(8, 3)\) ways.Case 3: If the fourth box has 5 balls, we then distribute 3 balls over 4 boxes with no other restrictions, which is \(C(3 + 4 - 1, 4 - 1) = C(6, 3)\) ways.Case 4: If the fourth box has 7 balls, we then distribute 1 ball over 4 boxes with no other restrictions, which is \(C(1 + 4 - 1, 4 - 1) = C(4, 3)\) ways.The total number of ways for part (b) is then the sum of the number of ways for each of these four cases.