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mathmath333

  • one year ago

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

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  1. anonymous
    • one year ago
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    Are toys equal ?

  2. mathmath333
    • one year ago
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    \(\large \color{black}{\begin{align}&a.)\ 2 \quad \quad &b.)\ 6 \hspace{.33em}\\~\\ &c.)\ 3 \quad \quad &d.)\ 4 \hspace{.33em}\\~\\ &e.)\ 5 \hspace{.33em}\\~\\ \end{align}}\)

  3. anonymous
    • one year ago
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    I mean same :)

  4. mathmath333
    • one year ago
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    that information is not given in the question

  5. mathmath333
    • one year ago
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    but from the context of the question i assume that the toys are same

  6. mathmath333
    • one year ago
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    @SithsAndGiggles

  7. anonymous
    • one year ago
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    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
    • one year ago
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    thnx

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