Question

There are four different types of bananas. In how many ways can 12 children select bananas so that at least one banana is selected from each type?

Hint:

When a problem says “at least one from each category”, subtract 1 from each variable first, then apply the Stars and Bars formula for non-negative solutions.

Answer 1

Correct Answer: 165

Let \( x_1, x_2, x_3, x_4 \) be the number of children selecting type 1, type 2, type 3 and type 4 bananas respectively.
Since total children are 12,
\[ x_1 + x_2 + x_3 + x_4 = 12 \] Step 1: Apply condition of at least one from each type.
Given that at least one banana of each type is selected,
\[ x_1 \ge 1,\; x_2 \ge 1,\; x_3 \ge 1,\; x_4 \ge 1 \] Let \[ y_i = x_i - 1 \] Then \[ y_1 + y_2 + y_3 + y_4 = 12 - 4 \] \[ = 8 \] where \[ y_i \ge 0 \] Step 2: Use Stars and Bars formula.
The number of non-negative integer solutions of \[ y_1 + y_2 + y_3 + y_4 = 8 \] is \[ \binom{8+4-1}{4-1} \] \[ = \binom{11}{3} \] Step 3: Calculate the value.
\[ \binom{11}{3} = \frac{11 \times 10 \times 9}{3 \times 2 \times 1} \] \[ = 165 \] Final Answer: \[ \boxed{165} \]