Question

If \(F(x, n)\) denotes the number of ways of distributing \(x\) toys to \(n\) children so that each child gets at most 2 toys, then what is the value of \(F(4, 3)\)?

• A. 2
• B. 3
• C. 4
• D. 6

Hint:

Split the 4 toys into groups no bigger than 2 per child, then count how many ways each group-shape can be handed to 3 different children.

Answer 1

The Correct Option is D

Step 1: Understand what \(F(4, 3)\) is asking.
We must hand out 4 identical toys to 3 children so that no child gets more than 2 toys. Since each child can hold at most 2, every child gets 0, 1 or 2 toys, and the three amounts must add up to 4.

Step 2: Find every way to split 4 into three numbers from \(\{0, 1, 2\}\).
Testing the possible combinations: \(2 + 2 + 0 = 4\) works, and \(2 + 1 + 1 = 4\) works. No other choice of three numbers from \(\{0, 1, 2\}\) adds to 4 (three 2's would give 6, and one 2 with two 0's only gives 2). So there are exactly two "shapes" the split can take: \((2, 2, 0)\) and \((2, 1, 1)\), spread over the 3 children.

Step 3: Count the arrangements for each shape.
For the shape \((2, 2, 0)\): we only need to choose which one of the 3 children gets 0 toys, since the other two automatically get 2 each. That gives 3 arrangements.
For the shape \((2, 1, 1)\): we only need to choose which one of the 3 children gets 2 toys, since the other two automatically get 1 each. That again gives 3 arrangements.

Step 4: Add the two cases together.
Total number of ways \(= 3 + 3 = 6\).
As a check, this matches the coefficient of \(x^4\) in \((1 + x + x^2)^3 = 1 + 3x + 6x^2 + 7x^3 + 6x^4 + 3x^5 + x^6\), which is the standard way to count "each of 3 recipients gets 0, 1 or 2 units, total 4 units": the coefficient of \(x^4\) is 6, confirming the case-by-case count.

Final Answer:
Options 2, 3 and 4 undercount because they only capture one of the two shapes or a fraction of the arrangements; only adding both shapes gives the full count.
\[ \boxed{F(4, 3) = 6} \]