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: Count all ways with no upper limit.
First ignore the "at most 2" rule. Distributing 4 identical toys among 3 children with no limit is a standard stars-and-bars count: the number of ways equals $\binom{4 + 3 - 1}{3 - 1} = \binom{6}{2} = 15$.

Step 2: Subtract the cases that break the "at most 2" rule.
A case breaks the rule if some child gets 3 or 4 toys. Let $A_i$ be the event that child $i$ gets at least 3 toys. If child $i$ gets at least 3, give that child 3 toys first, leaving only 1 toy to distribute freely among the 3 children: that is $\binom{1 + 3 - 1}{3 - 1} = \binom{3}{2} = 3$ ways. Since there are 3 children, the total count over all $A_i$ (allowing overlap) is $3 \times 3 = 9$.

Step 3: Check for double-counted cases.
Could two children each get 3 or more toys at once? That would need at least $3 + 3 = 6$ toys, but we only have 4, so no such case exists. There is nothing to add back, so the number of "bad" distributions is exactly $9$.

Step 4: Subtract to get the valid count.
Valid distributions $= 15 - 9 = 6$.
This matches the case-by-case count of the two shapes $(2,2,0)$ and $(2,1,1)$, confirming the answer through a completely different counting route.

Final Answer:
$F(4, 3) = 6$
\[ \boxed{6} \]