Part (a): Unrestricted choices Since there is no restriction, each child independently may choose any of the three flavors. Concretely: Child 1 has 3 choices ( V, C, or S ). Child 2 has 3 choices, and so on through child 5. By the multiplication principle, the total number of ways is $$ 3 \times 3 \times 3 \times 3 \times 3 = 3^5 = 243. $$ Part (b): Every flavor appears at least once Now we require that among the five cones handed out, vanilla, chocolate, and strawberry each be chosen by at least one child. Equivalently, no flavor is “unused.” We’ll use inclusion–exclusion to count exactly those assignments that include all three flavors. Count all assignments without restriction. As in part (a), there are $$ 3^5 = 243 $$ total assignments if we ignore the “use‐all‐three” requirement. Subtract assignments that omit exactly one flavor. Suppose vanilla “V” is never chosen. Then each of the 5 children must choose from just $\{C, S\}$. There are $$ 2^5 = 32 $$ ways in which every child’s choice is either C or S. Similarly, there are $$ 2^5 = 32 $$ assignments that avoid chocolate (all choices in $\{V, S\}$), and 32 that avoid strawberry (all choices in $\{V, C\}$). Since there are $$ \binom{3}{1} = 3 $$ ways to pick which single flavor is omitted, we subtract $$ 3 \times 2^5 = 3 \times 32 = 96. $$ Add back assignments that omit exactly two flavors. If two flavors—for instance, $\{V, C\}$—are both omitted, then every child must choose the remaining single flavor (strawberry S). That forces exactly $$ 1 $$ assignment (everyone picks S). There are $$ \binom{3}{2} = 3 $$ ways to choose which two flavors are omitted: $\{V, C\}, \{V, S\}, \{C, S\}$. For each pair, there is exactly $$ 1^5 = 1 $$ assignment (all kids pick the single leftover flavor). Because those “double‐omission” assignments were subtracted twice in step 2, we add them back once each. Thus we add $$ 3 \times 1 = 3. $$ Putting these together, the count of assignments that use all three flavors is $$ \underset{\text{all assignments}}{\underbrace{3^5}} - \underset{\text{omit 1 flavor}}{\underbrace{3 \cdot 2^5}} + \underset{\text{omit 2 flavors}}{\underbrace{3 \cdot 1^5}} = 243 - 96 + 3 = 150. $$ Part (c): General $n$ children, $k$ flavors We now have $n$ distinct children (labeled $1,\dots,n$) and $k$ distinct flavors (labeled $1,\dots,k$). Each child receives exactly one cone. We require that every flavor appears at least once. Count all assignments without restriction. Each child can choose any of the $k$ flavors, so there are $$ k^n $$ total assignments if we ignore the “use‐all‐$k$” requirement. Subtract assignments that omit exactly one flavor. If a fixed flavor (say flavor $i$) is never chosen, then each child must pick from the remaining $k - 1$ flavors, giving $$ (k - 1)^n $$ assignments. There are $\binom{k}{1}$ ways to choose which single flavor is omitted, so we subtract $$ \binom{k}{1} (k - 1)^n. $$ Add back assignments that omit exactly two flavors. If two flavors $\{i, j\}$ are both omitted, then each child picks from the remaining $k - 2$ flavors, giving $$ (k - 2)^n $$ assignments. There are $\binom{k}{2}$ ways to choose which two flavors are omitted, so we add back $$ \binom{k}{2} (k - 2)^n. $$ Continue alternating until omitting $k$ flavors (which contributes 0 if $n > 0$). Putting everything together by inclusion–exclusion, the number of assignments that use all $k$ flavors is $$ \sum_{j=0}^{k} (-1)^j \binom{k}{j} (k - j)^n. $$ Hence, $$ \boxed{ \sum_{j=0}^{k} (-1)^j \binom{k}{j} (k - j)^n } $$

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