Question

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

Answer 1

Are toys equal ?

Answer 2

\(\large \color{black}{\begin{align}&a.)\ 2 \quad \quad
&b.)\ 6 \hspace{.33em}\\~\\
&c.)\ 3 \quad \quad
&d.)\ 4 \hspace{.33em}\\~\\
&e.)\ 5 \hspace{.33em}\\~\\
\end{align}}\)

Answer 3

I mean same :)

Answer 4

that information is not given in the question

Answer 5

but from the context of the question i assume that the toys are same

Answer 6

@SithsAndGiggles

Answer 7

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:)

Answer 8

thnx