Lesson Math circle level problem on binomial coefficients

Math circle level problem on binomial coefficients

Problem 1

I will consider consequently the cases when where is no envelopes in the box #3;
                                        when there is 1 envelope  in the box #3;
                                        when there is 2 envelopes in the box #3;
                                        when there is 3 envelopes in the box #3;
                                        when there is 4 envelopes in the box #3;
                                        when there is 5 envelopes in the box #3.

For each of these cases, I will calculate the number of different distributions of the envelopes in three mailboxes, as follows.



a)  0 envelopes in the box #3:   =  +  +  +  +  +  = 1 + 5 + 10 + 10 + 5 + 1 = 32 ways.


b)  1 envelope in the box #3:   = 5*( +  +  +  + ) = 5*(1 + 4 + 6 + 4 + 1) = 80 ways.


c)  2 envelopes in the box #3:   = 10*( +  +  + ) = 10*(1 + 3 + 3 + 1) = 80.


d)  3 envelopes in the box #3:   = 10*( +  + ) = 10*(1 + 2 + 1) = 40 ways.


e)  4 envelopes in the box #3:   = 5*( + ) = 5*2 = 10 ways.


f)  5 envelopes in the box #3:   = 1 way.



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


Answer.  243 ways.

You need to consider the binomial expansion of   as the sum 

     = sum of all   with the coefficients  ,  i + j + k = 5.


Each particular term    with i + j + k = 5 "marks" some possible particular distribution 
of envelopes in the three boxes called "x", "y" and "z",
and the coefficient    at this term represents the "multiplicity" of the given concrete distribution 
(i,j,k) in the boxes x, y, and z.


The number of all possible distributions is the sum of all coefficients  , and it is equal to 

the value of     at  x= 1, y= 1, z= 1,  which is exactly   =  = 243.