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document.write( "a) 0 envelopes in the box #3: 
= 
+ 
+ 
+ 
+ + = 1 + 5 + 10 + 10 + 5 + 1 = 32 ways.\r\n" );
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document.write( "b) 1 envelope in the box #3: 
= 5*(
+ 
+ 
+ 
+ 
) = 5*(1 + 4 + 6 + 4 + 1) = 80 ways.\r\n" );
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document.write( "c) 2 envelopes in the box #3: 
= 10*(
+ 
+ 
+ 
) = 10*(1 + 3 + 3 + 1) = 80.\r\n" );
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document.write( "d) 3 envelopes in the box #3: 
= 10*(
+ 
+ 
) = 10*(1 + 2 + 1) = 40 ways.\r\n" );
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document.write( "e) 4 envelopes in the box #3: 
= 5*(
+ ) = 5*2 = 10 ways.\r\n" );
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document.write( "f) 5 envelopes in the box #3: 
= 1 way.\r\n" );
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document.write( "The total is the sum 32 + 80 + 80 + 40 + 10 + 1 = 243 ways.\r\n" );
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document.write( "Answer. 243 ways.\r\n" );
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document.write( "The formulas in this post are SELF-EXPLANATORY. \r
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document.write( "If you have questions or if you need explanations, look into the formulas until they tell you the whole story.\r
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document.write( "==================\r
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document.write( "Notice that 243 = 
, and it is not eventually.\r
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document.write( "There is another way to calculate it, which gives the same result, but is much shorter and much more elegant.\r
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document.write( "You need to consider the binomial expansion of 
as the sum \r\n" );
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document.write( " = sum of all 
with the coefficients 
, i + j + k = 5.\r\n" );
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document.write( "Each particular term 
with i + j + k = 5 \"marks\" each possible particular distribution \r\n" );
document.write( "of envelopes in three boxes called \"x\", \"y\" and \"z\".\r\n" );
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document.write( "The number of all possible distributions is the sum of all coefficients , and it is equal to \r\n" );
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document.write( "the value of at x= 1, y= 1, z= 1, which is exactly 
= = 243.\r\n" );
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document.write( "This problem is for an advanced Math circle / Math Olympiad level.\r
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document.write( "Therefore, I will not go further into details - the idea is just presented very clearly for an adequate person.\r
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