α + β + γ = 1 0 ∑ α ! β ! γ ! 1 0 !
Let α , β , γ be non-negative integers. Find the value of the summation above.
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Consider expanison of ( x + y + z ) 1 0 .
∑ α + β + γ = 1 0 α ! β ! γ ! 1 0 ! is equal to o the sum of the coefficients in the expansion , which we get by putiing x = y = z = 1
Therefore, α + β + γ = 1 0 ∑ α ! β ! γ ! 1 0 ! = 3 1 0 .
I think we are missing something.Alfa ,bete,gamma are given as positive.The answer 3^10 is true when they are non negative(including 0)
Same approach
It is simply the sum of the coefficients of a polynomial distribution that is equal 3 n for an n times repetition of a test having 3 distinct final states each time. Compare this please with the binomial distribution, there we get 2 n .
The problem needs to be fixed: the solution only works if the numbers are non-negative (including 0 ). If the numbers are positive, the answer is 5 5 9 8 0 .
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There are 10 objects. You color α of them red, β of them green, and γ of them blue; there are α ! β ! γ ! 1 0 ! ways to do this. Summing over all α , β , γ whose sum is 10 gives all possible colorings of the 10 objects with red, green, and blue. But we know that the number of such colorings is just 3 1 0 , since each object can be colored independently in three ways.