SOLUTION: In how many ways could five different envelopes be distributed into three mailboxes? The answer is 300 but I don’t understand how to do it please help

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Question 1125168: In how many ways could five different envelopes be distributed into three mailboxes? The answer is 300 but I don’t understand how to do it please help
Answer by ikleyn(52776) About Me  (Show Source):
You can put this solution on YOUR website!
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a)  0 envelopes in the box #3:  N%5B0%5D = C%5B5%5D%5E0 + C%5B5%5D%5E1 + C%5B5%5D%5E2 + C%5B5%5D%5E3 + C%5B5%5D%5E1 + C%5B5%5D%5E0 = 1 + 5 + 10 + 10 + 5 + 1 = 32 ways.


b)  1 envelope in the box #3:  N%5B1%5D = 5*(C%5B4%5D%5E0 + C%5B4%5D%5E1 + C%5B4%5D%5E2 + C%5B4%5D%5E3 + C%5B4%5D%5E4) = 5*(1 + 4 + 6 + 4 + 1) = 80 ways.


c)  2 envelopes in the box #3:  N%5B2%5D = 10*(C%5B3%5D%5E0 + C%5B3%5D%5E1 + C%5B3%5D%5E2 + C%5B3%5D%5E3) = 10*(1 + 3 + 3 + 1) = 80.


d)  3 envelopes in the box #3:  N%5B3%5D = 10*(C%5B2%5D%5E0 + C%5B2%5D%5E1 + C%5B2%5D%5E2) = 10*(1 + 2 + 1) = 40 ways.


e)  4 envelopes in the box #3:  N%5B4%5D = 5*(C%5B1%5D%5E0 + C%5B2%5D%5E1) = 5*2 = 10 ways.


f)  5 envelopes in the box #3:  N%5B5%5D = 1 way.



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


Answer.  243 ways.

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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   243 = 3%5E5,  and it is not eventually.

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


You need to consider the binomial expansion of  %28x%2By%2Bz%29%5E5 as the sum 

    %28x%2By%2Bz%29%5E5 = sum of all  A%5Bi%2Cj%2Ck%5D%2Ax%5Ei%2Ay%5Ej%2Az%5Ek with the coefficients  A%5Bi%2Cj%2Ck%5D,  i + j + k = 5.


Each particular term  x%5Ei%2Ay%5Ej%2Az%5Ek  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%5Bi%2Cj%2Ck%5D, and it is equal to 

the value of  %28x%2By%2Bz%29%5E5   at  x= 1, y= 1, z= 1,  which is exactly  %281%2B1%2B1%29%5E5 = 3%5E5 = 243.


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

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