## mathmath333 one year ago \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}}

1. anonymous

Are toys equal ?

2. mathmath333

\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}}

3. anonymous

I mean same :)

4. mathmath333

that information is not given in the question

5. mathmath333

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

6. mathmath333

@SithsAndGiggles

7. anonymous

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:)

8. mathmath333

thnx