Question 1125168
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<pre>
a)  0 envelopes in the box #3:  {{{N[0]}}} = {{{C[5]^0}}} + {{{C[5]^1}}} + {{{C[5]^2}}} + {{{C[5]^3}}} + {{{C[5]^1}}} + {{{C[5]^0}}} = 1 + 5 + 10 + 10 + 5 + 1 = 32 ways.


b)  1 envelope in the box #3:  {{{N[1]}}} = 5*({{{C[4]^0}}} + {{{C[4]^1}}} + {{{C[4]^2}}} + {{{C[4]^3}}} + {{{C[4]^4}}}) = 5*(1 + 4 + 6 + 4 + 1) = 80 ways.


c)  2 envelopes in the box #3:  {{{N[2]}}} = 10*({{{C[3]^0}}} + {{{C[3]^1}}} + {{{C[3]^2}}} + {{{C[3]^3}}}) = 10*(1 + 3 + 3 + 1) = 80.


d)  3 envelopes in the box #3:  {{{N[3]}}} = 10*({{{C[2]^0}}} + {{{C[2]^1}}} + {{{C[2]^2}}}) = 10*(1 + 2 + 1) = 40 ways.


e)  4 envelopes in the box #3:  {{{N[4]}}} = 5*({{{C[1]^0}}} + {{{C[2]^1}}}) = 5*2 = 10 ways.


f)  5 envelopes in the box #3:  {{{N[5]}}} = 1 way.



The total is the sum   32 + 80 + 80 + 40 + 10 + 1 = 243 ways.


<U>Answer</U>.  243 ways.
</pre>

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The formulas in this post are SELF-EXPLANATORY.  


If you have questions or if you need explanations, look into the formulas until they tell you the whole story.



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Notice that  &nbsp;&nbsp;243 = {{{3^5}}}, &nbsp;and it is not eventually.


There is another way to calculate it, &nbsp;which gives the same result, &nbsp;but is much shorter and much more elegant.



<pre>
You need to consider the binomial expansion of  {{{(x+y+z)^5}}} as the sum 

    {{{(x+y+z)^5}}} = sum of all  {{{A[i,j,k]*x^i*y^j*z^k}}} with the coefficients  {{{A[i,j,k]}}},  i + j + k = 5.


Each particular term  {{{x^i*y^j*z^k}}}  with i + j + k = 5 "marks" each possible particular distribution 
of envelopes in three boxes called "x", "y" and "z".


The number of all possible distributions is the sum of all coefficients  {{{A[i,j,k]}}}, and it is equal to 

the value of  {{{(x+y+z)^5}}}   at  x= 1, y= 1, z= 1,  which is exactly  {{{(1+1+1)^5}}} = {{{3^5}}} = 243.
</pre>


This problem is for an <U>advanced Math circle</U> / <U>Math Olympiad</U> level.


Therefore, &nbsp;I will not go further into details - the idea is just presented very clearly for an adequate person.