\(\large \color{black}{\begin{align}& \normalsize \text{If }\ F(x,n)\ \text{be the number of ways distributing }\ "x" \hspace{.33em}\\~\\ & \normalsize \text{toys to }\ "n" \ \text{children so that each child receives at most 2 toys , }\ \hspace{.33em}\\~\\ & \normalsize \text{then }\ F(4,3)=\underline{\quad \quad} \hspace{.33em}\\~\\ \end{align}}\)

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\(\large \color{black}{\begin{align}& \normalsize \text{If }\ F(x,n)\ \text{be the number of ways distributing }\ "x" \hspace{.33em}\\~\\ & \normalsize \text{toys to }\ "n" \ \text{children so that each child receives at most 2 toys , }\ \hspace{.33em}\\~\\ & \normalsize \text{then }\ F(4,3)=\underline{\quad \quad} \hspace{.33em}\\~\\ \end{align}}\)

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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Are toys equal ?
\(\large \color{black}{\begin{align}&a.)\ 2 \quad \quad &b.)\ 6 \hspace{.33em}\\~\\ &c.)\ 3 \quad \quad &d.)\ 4 \hspace{.33em}\\~\\ &e.)\ 5 \hspace{.33em}\\~\\ \end{align}}\)
I mean same :)

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Other answers:

that information is not given in the question
but from the context of the question i assume that the toys are same
It depends on whether they're same or not but suppose that they are same and we have 3 children and 4 toys and we want to distribute them.There is two possibilities 1-1-2 and 2-2-0 because it says each child receives at most 2 .It can't be 1-0-3 we know this.So moving on; A) 1-1-2 one child is lucky because he/she gets most :) you can select in 3 ways the lucky child but the mathematical expression is 3!/2! (Actually we put them in order 112 121 211) So we have 3. B)2-2-0 in that case two of them are having all toys.Select 2 children among 3 can be found by the same way 3!/2! It is also 3. The result is 3+3=6 The theory behind it is permutation with repetition.I hope it helps you:)
thnx

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